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Integration — 1/10
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Integration — 2/10
Example 1
Find:
Contents
1 Introduction
2
1.1 Indefinite Integration . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Definite Integration . . . . . . . . . . . . . . . . . . . . . . . 3
2
Integration by Substitution
4
3
Integration by Parts
5
4
Integration Leading to Logarithms
6
5
Integrating Partial Fractions
7
6 Integrating Trigonometric Functions
8
6.1 Trigonomtric substitution . . . . . . . . . . . . . . . . . . . 9
6.2 Other Integrals to be aware of . . . . . . . . . . . . . . . . 9
1. Introduction
Differentiation and integration are two interrelated areas of
mathematics and make up the branch of maths broadly called
calculus. You may recall that differentiating a function gives
the gradient of a curve. In other words, the derivative describes how the gradient of a function f (x) changes with x.
Integration can be thought of as the reverse process of differentiation. So if we find the derivative of a function and then
integrate the resulting derivative we return back to the original
function. Broadly then, the integral is the inverse function of
the derivative.
You may recall that differentiating a polynomial:
y = xn → f 0 (x) =
dy
= nxn−1
dx
Bearing in mind that integration is the reverse process of
differentiation if we are to integrate f (x) = xn :
Z
xn dx =
xn+1
+C
(n + 1)
Instead of bringing the power down and then reducing the
power by one, in integration we increase the power by one and
then divide by the increased power. Note the above integral
is a so-called indefinite integral. It is important to note the
constant C added to the evaluated integral. This constant
must be added when evaluating all indefinite integrals. The
reasoning behind why a constant is needed can be thought
of in the following way: if integration is the reverse process
of differentiation then when we integrate the derivative of
a function we return back to the original function. So an
unknown constant is added because the original function may
have had an additional constant that would be lost in the
process of differentiation.
1.1 Indefinite Integration
Indefinite integration formally involves finding the antiderivatives of a function f . The process of finding the antiderivative
is called integration. In other words, the result of finding the
integral of a function gives the function that when differentiated gives back the function we are integrating.
Z
x7 dx
The antiderivative of x7 is 18 x8 . Hence, this means the x7 dx =
1 8
8 x + C. We can check our final result by differentiating
f (x) = 18 x8 +C and the derivative f 0 (x) = x7 (differentiating
a constant C is equal to 0). Thus it is quite easy to check the
answer. If you take the derivative of the final result it should
equal the original function (i.e. the integrand).
Example 2
Find:
R
Z
(4x3 + 6x2 − 7x + 3) dx
Now finding the integral or the antiderivative:
7
(4x3 + 6x2 − 7x + 3) dx = x4 + 2x3 − x2 + 3x +C
2
Example 3
Find:
Z
Z
x2 − 5x + 2
dx =
x
Z
2
x − 5 + dx
x
Rewriting 2/x as 2x−1 we can then carry out the integral:
Z
1
x − 5 + 2x−1 dx = x2 − 5x + 2 ln x +C
2
Here we’re using the rule that 1x dx = ln |x| +C.
Example 4
Find:
Z
√
√
4 3
3
4 x − 2 − + 2 + 3 x2 dx
x
x
First we rewrite the integrand in terms of fractional powers
before we can carry out the integral:
R
Z
4x1/2 − 4x−2 − 3x−1 + 2 + 3x2/3 dx
Thus the evaluated integral then becomes:
Z
4x1/2 −4x−2 −3x−1 +2+3x2/3 dx = 4
+2x + 3
x5/3
8
9
= x3/2 + 4x−1 − 3 ln |x| + 2x + x5/3
5/3
3
5
Examples
Z
x9 dx
Z
x1/2 dx
Z
1
dx
x4
Z
e4x dx
1)
2)
3)
4)
x3/2
x−1
−4
−3 ln |x|
3/2
−1
Integration — 3/10
1.2 Definite Integration
Also integration gives the area beneath the curve. Definite
integration involves finding the area under a curve where the
integration has limits. For example, to integrate the function
y = x2 between x = 1 and x = 3 would be written as:
Z 3
Z 3
A=
y dx
0
Z 3
=
x dx
1
Then carrying out the integration:
3
Z 3
1
x2 dx = x3
3
1
1
We then substitute in the upper and lower limits and then take
the difference:
Z 3
1 3 3
1 3
1 3
27 1 26
2
x dx = x
= 3 − 1 =
− =
3
3
3
3
3
3
1
1
This then represents the area beneath the curve of the function
y = x2 , between the lines x = 1 and x = 3.
Example 1
Find the area between the curve y = x(x − 3) and the coordinates x = 0 and x = 5. First it is a good idea to sketch the
curve to get an idea of the region bounded by x = 0 and x = 5.
If we set the function equal to 0:
=
x3 3x2
−
3
2
3
0
27 (3 × 9)
0 (3 × 0)
=
−
−
−
3
2
3
2
27
= 9−
− [0]
2
1
2
Note that in all cases when the area is bounded below the
x-axis the area is always negative. So we take the absolute
value of the found area.
Now we need to find the area B between the curve and the
x-axis between the coordinates x = 3 and x = 5:
= −4
Z 5
B=
y dx
3
y = x(x − 3) = 0
Z 5
we can find the x-coordinates where the function intercepts
the x-axis. Thus, the function intercepts the x-axis at x = 0
and x = 3. A sketch of the curve is shown below with the two
bounded areas A and B:
(x2 − 3x) dx
0
2
=
(x2 − 3x) dx
3
=
=
x3 3x2
−
3
2
5
3
27 (3 × 9)
125 (3 × 25)
−
−
−
3
2
3
2
2
1
1
= 41 − 37 − 9 + 13
3
2
2
2
=8
3
So the total area beneath the curve is 4 21 + 8 23 = 13 61 .
Examples
Evaluate the following integrals:
Z 1
Figure 1. Pot of the function y = x(x − 3) with the areas A
and B delineated. The area B above the axis is positive and
the area A below the axis is negative. In general all areas
bounded above the axis are positive and all areas bounded
below the axis are negative.
1)
2)
Z 3
1
2
Z 1
3)
First we find the area A, between the curve and the x-axis
coordinates x = 0 and x = 3. As the area is totally below the
x-axis we expect the calculated area to be negative:
x2 dx
0
x2
dx
e2x dx
0
Z π/2
4)
sin(x) dx
0
Integration — 4/10
2. Integration by Substitution
Similar to differentiating using substitution, integration by
substitution involves simplifying an integral by setting it equal
to another variable name. For example, it is difficult to complete the integral:
Z
(2x + 2)ex
2 +2x+3
but now since we have the integral in terms of u we need to
change the variable over which we are integrating over from
x to u. Using the derivative of u, we can then rearrange the
derivative and write dx = du/(2x + 2). The integral we can
then write as:
du
(2x + 2)e
(2x + 2)
Z
u
Example 2
Integrate:
u
(2x + 2)e dx =
1 1
u 2 du
7
Carrying out the integral and substituting back for u in terms
of x:
1
1
u7 du = u8 +C = (x2 + 5x)8 +C
8
8
Z
Example 3
Integrate:
√
sin( x)
√
dx
x
Z
So the integral just becomes:
Z
u
u
e du = e +C
Now we can substitute back for u in the integral and we have
evaluated the integral:
Z
eu du = eu +C = ex
2 +2x+3
+C
√
We then choose u such that u = x and then differentiating
1
with respect to x the derivative becomes du/dx = 12 x− 2 . Then
rearranging for dx:
√
dx = 2 xdu
Then, substituting for u and dx in the original integral:
Z
So now we have fully evaluated the integral. We were able
to evaluate the integral by simplifying the original integral
and rearranging the derivative of u and substituting for dx. It
must be stressed that integration by substitution is a skill that
gets easier with practice. By doing more and more integrals
it becomes easier to spot which part of the integral should be
set equal to u. Sometimes the integral is complex and it is
difficult to spot what to substitute in the integral and you may
need to try a variety of substitutions until you are left with an
integral you are able to evaluate.
2
√
sin(u)du = −2 cos(u) +C = −2 cos( x) +C
Where we have substituted back for u in the evaluated integral
to put the integral back in terms of x.
Examples
Integrate the following by making the appropriate substitution:
Z
1)
Z
Example 1
Integrate:
(7x + 9) dx
We choose u such that u = 7x + 9 and then differentiating
with respect to x the derivative becomes du/dx = 7. Then
substituting in for u and dx:
Z
(2x + 2)eu dx
Z
1
1
u7 du = u8 +C = (x2 + 5x)8 +C
8
8
Z
Z p
dx
However, by setting u = x2 + 2x + 3 and then differentiating,
du/dx = 2x + 2. We can then rewrite the integral as:
Z
and then replacing dx with du/(2x + 5). The integral becomes:
2)
(2x − 1)7 dx
sin(7x − 3) dx
Z
Z
2
7
(2x + 5)(x + 5x) dx
Then setting u = x2 + 5x and this means that du/dx = 2x + 5.
Then, substituting u into the integral we are left with:
Z
(2x + 5)u7 dx
3)
2x
Z
1
dx
1 − 2x
Z
x sin(2x2 ) dx
4)
5)
p
1 + x2 dx
Integration — 5/10
3. Integration by Parts
As discussed in the previous section, differentiation and integration are closely linked. When we differentiate a function
to find the gradient, by integrating the derivative we return to
the original function. An important and commonly used rule
in differentiation is the product rule. Recall that the product
rule of differentiation allows one to differentiate a product of
two functions of x, which we usually write as u(x) and v(x):
du(x)
dv(x)
dy(x)
= v(x)
+ u(x)
dx
dx
dx
y0 (x) = v(x)u0 (x) + u(x)v0 (x)
Z
Z
uv dx =
v
du
dx +
dx
Z
u
dv
dx =
dx
Z
Z
uv dx =
du
v dx = uv −
dx
Z
d(uv)
dx
dx
dv
u dx
dx
x
Z
xe dx =
du
v dx = uv −
dx
Z
Z
(2)
dv
u dx
dx
Here we set v = x and du/dx = ex so this means dv/dx = 1
and u = ex . Thus, we now have everything we need to use
the integration by parts by formula. By substituting into the
integration by parts formula:
Z
xex dx =
Z
xex dx = xex −
Z
u=x
and
Now we can substitute into the integration by parts formulae
(Equation 2) and carry out the final integration:
Therefore, if we are required to integrate the product of two
functions u and v we use the formula derived above. The best
way to get familiar with this formula is to demonstrate its use
by example. Consider the integral:
Z
Here it is not obvious we are integrating a product of two
functions but we can write ln x as 1 · ln x. Then, with use
of LATE, we differentiate the ln x function and integrate 1.
Therefore, v = ln x and du/dx = 1. This means:
dv 1
=
dx x
which we can write as:
Z
ln x dx
(1)
We can use the product rule of differentiation to derive the
equivalent rule for integrating a product of two functions of
x, called integration by parts. If we integrate both sides of
Equation 1 we get:
Z
we differentiate (i.e. set equal to v in the formula) and likewise
to decide which function to integrate (i.e. set equal to du/dx)
we work from the end of LATE. Thus, for the example we
just went through we find out what to differentiate by working through L→A→T→E. So, as we move to the right, we
find Algebra and as we move to the left we find Exponentials.
Thus we differentiate the algebraic term and integrate the exponential term-exactly what was done in the example.
Example 1
Integrate:
ln x dx = x ln x −
Z
1
x · dx = x ln x − x +C
x
Example 2
Integrate:
Z
x sin x dx
Using the same ideas as before to decide on the terms to
differentiate and integrate, we set v = x and du/dx = sin x.
Carrying out the integration and differentiation:
dv
=1
dx
and
u = − cos x
Substituting in the appropriate terms into the integration by
parts formulae and carrying out the integration:
ex dx = xex − ex +C
Z
In this example it may seem arbitrary what we chose for v
and du/dx. However, it is important to pick carefully in order
that it is possible to do the integral. Similar to SOHCAHTOA
used to remember the definitions of sin, cos and tan, we can
use LATE to correctly decide which term we should differentiate and integrate in the integration by parts formula. In
the LATE pneumonic: Logarithms, Algebra, Trigonometry
and Exponentials. To find out what we should differentiate in the product we begin at the beginning of LATE with
L→A→T→E working through the word until we find one of
the two functions in the product to be integrated. The first
function we encounter as we move to the right is the function
x sin x dx = x(− cos x) −
= −x cos x +
Z
(− cos x) dx
Z
cos x dx
= −x cos x + sin x +C
Example 3
Integrate:
Z
x cos 3x dx
Similar to Example 2, here we set v = x and du/dx = cos 3x.
Differentiating and integrating:
Integration — 6/10
4. Integration Leading to Logarithms
dv
=1
dx
and
u=
1
sin 3x
3
If the integrand is in a particular form evaluating the integral
leads to a logarithm function:
Therefore:
Z
1
1
x cos 3x dx = x sin 3x −
3
3
Z
Z
sin 3x dx
Then carrying out the final integration:
1
1
x sin 3x −
3
3
Z
1
1
sin 3x dx = x sin 3x + cos 3x +C
3
9
Note, in all the worked examples above it was only required
to use the integration by parts formula once to evaluate the
integral. However, there are cases where you may have to use
the integration by parts formula twice to completely evaluate
the integral i.e. on the first use of the integration by parts
formula you are left with an integral that you still cannot evaluate and so need to use the integration by parts formula again.
Based on the previous examples: Integrate the following by
integrating by parts twice:
Z
i.e. when the numerator is the differential of the denominator
we simply take the natural logarithm of the denominator. However, it is not always the case that the numerator is the exact
differential of the denominator. In these cases the integrand
may have to be manipulated to arrive at the form f 0 (x)/ f (x).
For example, in a simple case we may have the integral:
Z
1
dx
2x
However, note that the numerator is not the exact differential
of the denominator. This means we need to multiply the
numerator by 2 and the denominator by 2 in order not to
change the overall integral. This can be written as:
x2 e3x dx
1
2
Answer:
1 2 3x 2 3x 2 3x
x e − xe + e +C
3
9
27
Examples
Integrate the following by using the integration by parts formula. In example 5, you will have to integrate by using the
integration by parts formula twice:
Z
1)
x ln x dx
Z
2)
ln x
dx
x5
Z
xe6x dx
Z
x
(3x + 5) cos dx
5
Z
(ln x)2 dx
3)
4)
5)
f 0 (x)
dx = ln |x| +C
f (x)
Z
2
1
dx = ln |2x| +C
2x
2
Note, we haven’t changed the overall integral and we have
taken the logarithm of the denominator.
Example 1
Integrate:
Z
tan(x) dx
First we begin by rewriting tan(x) = sin(x)/ cos(x) and then
to get the integrand in the form f 0 (x)/ f (x), sin(x) needs to
be multiplied by −1 so we must divide by −1 to leave the
integrand unchanged:
−
Z
− sin(x)
dx = − ln(cos x) +C
cos(x)
Example 2
Integrate:
e2x
Z
e2x − 1
dx
The denominator e2x − 1 when differentiated leads to 2e2x ,
very similar to the numerator of e2x . To manipulate the integrand such that the numerator is the exact differential of
the denominator we need to multiply and divide by 2. The
integral then becomes:
1
2
Z
2e2x
dx
e2x − 1
Integration — 7/10
Now we have the integral in the form where the numerator
is the exact differential of the denominator. This means we
can take the logarithm of the denominator. Evaluating the
integral:
1
2
2e2x
Z
e2x − 1
dx =
x cos x + sin x
dx
x sin x
First of all we need to think what we would obtain if we are
to differentiate the denominator. Using the product rule of
differentiation setting u = x and v = sin x this means du/dx = 1
and dv/dx = cos x. Thus the differential is:
dy
= sin x + x cos x
dx
Thus, using the product rule to differentiate the denominator
we find that the numerator is the exact differential of the denominator. Thus we can now go about evaluating the integral:
Z
Using the ideas discussed in Section 4, we can integrate more
complicated functions. For example, suppose we want to
integrate:
1
ln(e2x − 1) +C
2
Example 3
Integrate:
Z
5. Integrating Partial Fractions
x cos x + sin x
dx = ln(x sin x) +C
x sin x
we cannot integrate this function using any of the techniques
we have learned so far. To evaluate an integrand of this form
first we need to write the integrand in terms of partial fractions:
2x − 5
A
B
≡
+
(4x − 1)(x + 2) (4x − 1) (x + 2)
Then we need to solve for both A and B. First we multiply
both sides of the equation by (4x − 1)(x + 2) and we are left
with:
2x − 5 ≡ A(x + 2) + B(4x − 1)
To solve for A and B we substitute values of x that eliminates
either A or B. To solve first for A we substitute x = 1/4:
2·
1
9 9A
1
− 5 ≡ A( + 2) = − =
→ A = −2
4
4
2
4
Then to solve for B we substitute x = −2:
Examples
Integrate the following using the rule that the integral of a quotient, where the numerator is the derivative of the denominator,
is equal to the logarithm of the denominator:
2x − 5
dx
(4x − 1)(x + 2)
Z
−9 = −9B → B = 1
Thus, now we can rewrite the integrand in terms of partial
fractions:
Z
1)
cot x dx
Z
Z
2)
3
dx
2 + 3x
5e5x
3)
dx
e5x
Z
x
4)
dx
1 + 2x2
Z
1
5)
dx
x ln(x)
2x − 5
dx ≡
(4x − 1)(x + 2)
Z
−2
1
+
dx
(4x − 1) (x + 2)
Integrating separately then:
Z
Z
1
dx = ln |x + 2| +C
(x + 2)
and
−
Z
2
1
dx = −
(4x − 1)
2
Z
1
4
dx = − ln |4x − 1| +C
(4x − 1)
2
Note how in the second case we had to rearrange the integrand
to get it in the form f 0 (x)/ f (x) by multiplying and dividing
by 2. Thus, our final answer is:
Z
2x − 5
1
dx = ln |x + 2| − ln |4x − 1| +C
(4x − 1)(x + 2)
2
Integration — 8/10
Examples
1 + tan2 A = sec2 A
1 + cot2 A = csc2 A
Integrate the following by expressing the integrand
in terms of
R 0
partial fractions and then integrate using the rule f (x)/ f (x)dx = Example 1
ln |x| +C :
Integrate:
Z
3
dx
x(x + 1)
Z
1
dx
(x + 2)(x + 1)
Z
x
dx
(2x + 3)(x − 4)
Z
3x + 2
dx
(x − 1)(x + 7)
1)
2)
3)
4)
Z
Firstly, we need to use a trigonometric identity to substitute
for cos2 x. The identity we will use in this example will be:
cos 2A = 2 cos2 A − 1
Substituting then for cos2 x we are left with the integral:
1
2
6. Integrating Trigonometric Functions
It is quite common to come across integrating trigonometric
functions. If you remember how to differentiate sin x, cos x
and tan x then it is quite simple to work out how to integrate
trigonometric functions:
y = sin x;
dy
= cos x →
dx
cos2 xdx
Z
cos x = sin x +C
dy
= − sin x → − sin x = cos x +C
dx
Z
dy
y = tan x;
= sec2 x → sec2 x = tan x +C
dx
So, if we remember the derivatives of the trig functions we
can easily work out the integrals. However, you must be very
careful not to confuse minus signs. For example, the integral
of − sin x is cos x and the integral of sin x is − cos x. Thus, take
care to keep track of minus signs when finding the integral
from the known derivatives.
However, integrating the simple trig functions aren’t the
only integrals you will come across. You will come across
multiples of trigonometric functions, trig functions raised to
a power, the addition of trig functions etc. Most you will not
be able to integrate without making a substitution to break it
down into an integrateable form:
Z
y = cos x;
2 sin A cos B = sin(A + B) + sin(A − B)
2 cos A cos B = cos(A − B) + cos(A + B)
Z
(1 + cos 2x)dx
Now we can integrate as normal. Evaluating the integral:
1
2
Z
1
1
(1 + cos 2x)dx = x + sin 2x +C
2
4
Example 2
Integrate:
Z
sin(2x) cos(2x)dx
We need to substitute for the integrand in order to carry out
the integral. The identity we will use will be:
2 sin A cos B = sin(A + B) + sin(A − B)
Substituting for A = B = 2x we are left with:
2 sin(2x) cos(2x) = sin(4x)
Then, substituting for the integrand in the original integral we
are left with:
1
2
Z
1
sin(4x)dx = − cos(4x) +C
8
Example 3
Integrate:
Z
2 cos(5x) cos(3x)dx
We need to substitute for the integrand in order to carry out
the integral. The identity we will use will be:
2 cos A cos B = cos(A − B) + cos(A + B)
2 sin A sin B = cos(A − B) − cos(A − B)
sin2 A + cos2 B = 1
cos 2A = cos2 A − sin2 A
cos 2A = 2 cos2 A − 1
cos 2A = 1 − 2 sin2 A
Substituting then for A = 5x and B = 3x we are left with:
2 cos(5x) cos(3x) = cos(2x) + cos(8x)
Then, substituting for the integrand in the original integral we
are left with:
Z
sin 2A = 2 sin A cos A
cos(2x) + cos(8x)dx =
1
1
sin(2x) + sin(8x) +C
2
8
Integration — 9/10
More difficult examples can involve rearranging the integrand
into a form where a trigonometric substitution can be made
and also it may be necessary to integrate by substitution. The
following example is a difficult example that demonstrates
both techniques:
Example 4
Integrate:
Z
sin4 (x) cos3 (x)dx
First we rewrite the integral as:
Z
sin4 (x)(cos2 (x) · cos(x))dx
Now we can substitute for cos2 (x) using the identity that
cos2 x = 1 − sin2 x. The integral then becomes:
Z
1
√
dx
2
a − x2
We base our choice of the substitution in this case by noticing
that the denominator a2 − x2 is similar in form to the trigonometric identity 1 − sin2 A = cos2 A. Thus to get it into a form
that matches the denominator of the integral (a2 − a2 sin2 A =
a2 cos2 A). So we try the substitution x = a sin θ , meaning
x2 = a2 sin2 θ then dx/dθ = a cos θ and dx = a cos θ dθ . The
integral then becomes:
Z
Z
1
√
dx =
a2 − x2
u5 u7
(u − u )du =
− +C
5
7
4
6
1
Z
p
a2 − a2 sin2 θ
Z
1
√
a cos θ dθ
2
a cos2 θ
Z
1
a cos θ dθ
a cos θ
=
(sin4 x cos x − sin6 x cos x)dx
Now at this point we can make a substitution by writing
u = sin x and so du/dx = cos x. So substituting for u and
replacing dx = du/ cos x we eliminate the factor of cos x and
the integral becomes:
Z
6.2 Other Integrals to be aware of
Suppose we are asked to find:
=
a cos θ dθ
Z
=
1 dθ = θ +C
Now we substitute for θ by rearranging our substitution x =
a sin θ for θ itself. Thus, θ = sin−1 ax . So our integral:
Z
6.1 Trigonomtric substitution
If we are asked to find the integral:
x
1
√
dx = sin−1 +C
2
2
a
a −x
Examples
Z
1
dx
1 + x2
Now if we set x = tan θ . But you might ask, why would
we choose to set x = tan θ . This is because the denominator
reduces to 1 + tan2 θ which we know from the trigonometric
identity is equal to sec2 θ (sec2 θ = 1 + tan2 θ ). Now we
need to change the variable of the integral from dx to dθ by
differentiating x = tan θ which gives dx/dθ = sec2 θ . Thus,
dx = sec2 θ dθ . Now, rewriting the integral:
Z
1
× sec2 θ dθ
sec2 θ
Z
=
1 dθ = θ +C
= tan−1 x +C
This is an important standard result that can be generalised:
Z
1
1
x
dx = tan−1 +C
2
2
a +x
a
a
This is a standard result that is worth being aware of and is
useful to be able to derive or look up where necessary.
Using a trigonometric substitution where indicated compute
the following integrals:
x2
√
dx
16 − x2
Substitute x = 4 sin θ
Z
1
2)
dx
1 + 4x2
Z
1)
Substitute x = 12 tan θ
Z
3)
sin2 x cos2 x dx
Hint: use the double angle formula sin2 x =
2x
cos2 x = 1+cos
2
Z
4)
1−cos 2x
2
and
sin4 x cos2 x dx
Hint: use the double angle formula twice to evaluate the
integral
Z π/3
5)
2 cos 5x cos 3x dx
π/6
Hint: use the product formulae to evaluate the integral
Integration — 10/10
Answers
Integration Leading to Logarithms
Indefinite Integration
1) ln | sin x| +C
1)
2)
x10
+C
10
2x3/2
+C
3
1
3) − x−3 +C
3
4)
e4x
+C
4
2) ln |2 + 3x| +C
3) ln |e5x | = 5x +C
1
4) ln |1 + 2x2 | +C
4
5) ln | ln |x|| +C
Integrating Partial Fractions
Z
Definite Integration
1
3
2)
1
6
3)
e2 1
− = 3.195
2 2
4
3
ln |2x + 3| + ln |x − 4| +C
22
11
19
5
4) ln |x − 1| + ln |x + 7| +C
8
8
Integrating Trigonometric Functions
1
(2x − 1)7 +C
16
1
2) − cos(3 − 7x) +C
7
3
2
3) (x2 + 1) 2
3
1
4) − ln(1 − 2x) +C
2
1
5) − cos(2x2 ) +C
4
Integration by Parts
3)
3
3
−
dx
x x+1
3)
Integration by Substitution
x2
x2
ln |x| − +C
2
4
2) −
Z
2) ln |x + 1| − ln |x + 2| +C
4) 1
1)
3
dx =
x(x + 1)
= 3 ln |x| − 3 ln |x + 1| +C
1)
1)
1)
4 ln |x| + 1
+C
16x4
1 6x
e (6x − 1) +C
36
4) 5((3x + 5) sin(x/5) + 15 cos(x/5)) +C
5) x(ln2 x − 2 ln x + 2) +C
1 p
x
1) − x 16 − x2 + 8 sin−1 +C
2
4
1
2) tan−1 2x +C
2
1
1
1
3) − sin x cos3 x + cos x sin x + x +C
4
8
8
1
1 3
4) − sin x cos3 x − sin x cos3 x +C
6
8
1
1
+ cos x sin x + x +C
16
16
√
3
5)
= 2.165
8