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Standard Table Table Annette Pilkington Strategy for Integration Standard Table Problem Solving approach Faced with an integral, we must use a problem solving approach to finding the right method or combination of methods to apply. Annette Pilkington Strategy for Integration Standard Table Problem Solving approach Faced with an integral, we must use a problem solving approach to finding the right method or combination of methods to apply. I It may be possible to Simplify the integral e.g. Z Z cos x cot xdx = dx. sin x Annette Pilkington Strategy for Integration Standard Table Problem Solving approach Faced with an integral, we must use a problem solving approach to finding the right method or combination of methods to apply. I It may be possible to Simplify the integral e.g. Z Z cos x cot xdx = dx. sin x I It may be possible to simplify or solve the integral with a substitution e.g. Z 1 dx x(ln x)10 Annette Pilkington Strategy for Integration Standard Table Problem Solving approach Faced with an integral, we must use a problem solving approach to finding the right method or combination of methods to apply. I It may be possible to Simplify the integral e.g. Z Z cos x cot xdx = dx. sin x I It may be possible to simplify or solve the integral with a substitution e.g. Z 1 dx x(ln x)10 I if it is of the form Z sinn x cosm xdx, Z tann x secm xdx Z sin(nx) cos(mx)dx we can deal with it using the standard methods for trigonometric functions we have studied. Annette Pilkington Strategy for Integration Standard Table Problem Solving approach Faced with an integral, we must use a problem solving approach to finding the right method or combination of methods to apply. I It may be possible to Simplify the integral e.g. Z Z cos x cot xdx = dx. sin x I It may be possible to simplify or solve the integral with a substitution e.g. Z 1 dx x(ln x)10 I if it is of the form Z sinn x cosm xdx, Z tann x secm xdx Z sin(nx) cos(mx)dx we can deal with it using the standard methods for trigonometric functions we have studied. I If we are trying to integrate a rational function, we apply the techniques of the previous section. Annette Pilkington Strategy for Integration Standard Table Problem Solving approach Faced with an integral, we must use a problem solving approach to finding the right method or combination of methods to apply. I It may be possible to Simplify the integral e.g. Z Z cos x cot xdx = dx. sin x I It may be possible to simplify or solve the integral with a substitution e.g. Z 1 dx x(ln x)10 I if it is of the form Z sinn x cosm xdx, Z tann x secm xdx Z sin(nx) cos(mx)dx we can deal with it using the standard methods for trigonometric functions we have studied. I If we are trying to integrate a rational function, we apply the techniques of the previous section. I We should check if integration by parts will work. Annette Pilkington Strategy for Integration Standard Table Problem Solving approach I √ If the integral contains an expression of the form ±x 2 ± a2 we can use a trigonometric substitution. If the integral contains an expression of the √ form√n ax + b, the function may become a rational function when we use u = n ax + b, a rationalizing substitution. This may also p p work for integrals with expressions of the form n g (x) with u = n g (x) Annette Pilkington Strategy for Integration Standard Table Problem Solving approach I I √ If the integral contains an expression of the form ±x 2 ± a2 we can use a trigonometric substitution. If the integral contains an expression of the √ form√n ax + b, the function may become a rational function when we use u = n ax + b, a rationalizing substitution. This may also p p work for integrals with expressions of the form n g (x) with u = n g (x) You may be able to manipulate the integrand to change its form. e.g. Z Z sec x(sec x + tan x) sec xdx = sec x + tan x Annette Pilkington Strategy for Integration Standard Table Problem Solving approach I √ If the integral contains an expression of the form ±x 2 ± a2 we can use a trigonometric substitution. If the integral contains an expression of the √ form√n ax + b, the function may become a rational function when we use u = n ax + b, a rationalizing substitution. This may also p p work for integrals with expressions of the form n g (x) with u = n g (x) I You may be able to manipulate the integrand to change its form. e.g. Z Z sec x(sec x + tan x) sec xdx = sec x + tan x I The integral may resemble something you have already seen and you may see that a change of format or substitution will convert the integral to some basic integral that you have already worked through e.g. Z sin x cos xe sin x dx Annette Pilkington Strategy for Integration Standard Table Problem Solving approach I √ If the integral contains an expression of the form ±x 2 ± a2 we can use a trigonometric substitution. If the integral contains an expression of the √ form√n ax + b, the function may become a rational function when we use u = n ax + b, a rationalizing substitution. This may also p p work for integrals with expressions of the form n g (x) with u = n g (x) I You may be able to manipulate the integrand to change its form. e.g. Z Z sec x(sec x + tan x) sec xdx = sec x + tan x I The integral may resemble something you have already seen and you may see that a change of format or substitution will convert the integral to some basic integral that you have already worked through e.g. Z sin x cos xe sin x dx I Your solution may involve several steps. Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I ln x dx Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I ln x dx I Integration by parts ; let u = ln x, dv = dx ...... Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I ln x dx I Integration by parts ; let u = ln x, dv = dx ...... R I tan x dx Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I ln x dx I Integration by parts ; let u = ln x, dv = dx ...... R I tan x dx R R sin x I write as dx, let u = cos x...... tan x dx = cos x Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I ln x dx I Integration by parts ; let u = ln x, dv = dx ...... R I tan x dx R R sin x I write as dx, let u = cos x...... tan x dx = cos x R I sin3 x cos x dx Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I ln x dx I Integration by parts ; let u = ln x, dv = dx ...... R I tan x dx R R sin x I write as dx, let u = cos x...... tan x dx = cos x R I sin3 x cos x dx I Substitution, let u = sin x...... Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I ln x dx I Integration by parts ; let u = ln x, dv = dx ...... R I tan x dx R R sin x I write as dx, let u = cos x...... tan x dx = cos x R I sin3 x cos x dx I Substitution, let u = sin x...... R I √ 1 dx 2 25−x Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I ln x dx I Integration by parts ; let u = ln x, dv = dx ...... R I tan x dx R R sin x I write as dx, let u = cos x...... tan x dx = cos x R I sin3 x cos x dx I Substitution, let u = sin x...... R I √ 1 dx 2 25−x I Trig. substitution, x = 5 sin θ, − π2 ≤ θ ≤ Annette Pilkington π , 2 R √ Strategy for Integration 1 25−x 2 dx = sin−1 “ ” x 5 Standard Table Review Outline How you would approach the following integrals: R I ln x dx I Integration by parts ; let u = ln x, dv = dx ...... R I tan x dx R R sin x I write as dx, let u = cos x...... tan x dx = cos x R I sin3 x cos x dx I Substitution, let u = sin x...... R I √ 1 dx 2 25−x I I Trig. substitution, x = 5 sin θ, − π2 ≤ θ ≤ R sec x dx Annette Pilkington π , 2 R √ Strategy for Integration 1 25−x 2 dx = sin−1 “ ” x 5 Standard Table Review Outline How you would approach the following integrals: R I ln x dx I Integration by parts ; let u = ln x, dv = dx ...... R I tan x dx R R sin x I write as dx, let u = cos x...... tan x dx = cos x R I sin3 x cos x dx I Substitution, let u = sin x...... R I √ 1 dx 2 25−x I I I “ ” R −1 x √ 1 Trig. substitution, x = 5 sin θ, − π2 ≤ θ ≤ π2 , dx = sin 5 25−x 2 R sec x dx R R x(sec x+tan x) sec xdx = sec sec Let u = sec x + tan x. x+tan x Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I ln x dx I Integration by parts ; let u = ln x, dv = dx ...... R I tan x dx R R sin x I write as dx, let u = cos x...... tan x dx = cos x R I sin3 x cos x dx I Substitution, let u = sin x...... R I √ 1 dx 2 25−x I I I I “ ” R −1 x √ 1 Trig. substitution, x = 5 sin θ, − π2 ≤ θ ≤ π2 , dx = sin 5 25−x 2 R sec x dx R R x(sec x+tan x) sec xdx = sec sec Let u = sec x + tan x. x+tan x R √x e dx Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I ln x dx I Integration by parts ; let u = ln x, dv = dx ...... R I tan x dx R R sin x I write as dx, let u = cos x...... tan x dx = cos x R I sin3 x cos x dx I Substitution, let u = sin x...... R I √ 1 dx 2 25−x I I I I I “ ” R −1 x √ 1 Trig. substitution, x = 5 sin θ, − π2 ≤ θ ≤ π2 , dx = sin 5 25−x 2 R sec x dx R R x(sec x+tan x) sec xdx = sec sec Let u = sec x + tan x. x+tan x R √x e dx √ Try the substitution w = x Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I ln x dx I Integration by parts ; let u = ln x, dv = dx ...... R I tan x dx R R sin x I write as dx, let u = cos x...... tan x dx = cos x R I sin3 x cos x dx I Substitution, let u = sin x...... R I √ 1 dx 2 25−x I I I I I I “ ” R −1 x √ 1 Trig. substitution, x = 5 sin θ, − π2 ≤ θ ≤ π2 , dx = sin 5 25−x 2 R sec x dx R R x(sec x+tan x) sec xdx = sec sec Let u = sec x + tan x. x+tan x R √x e dx √ Try the substitution w = x 1 dw = 2√1 x dx = 2w dx, dx = 2w dw Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I ln x dx I Integration by parts ; let u = ln x, dv = dx ...... R I tan x dx R R sin x I write as dx, let u = cos x...... tan x dx = cos x R I sin3 x cos x dx I Substitution, let u = sin x...... R I √ 1 dx 2 25−x I I I I I I I “ ” R −1 x √ 1 Trig. substitution, x = 5 sin θ, − π2 ≤ θ ≤ π2 , dx = sin 5 25−x 2 R sec x dx R R x(sec x+tan x) sec xdx = sec sec Let u = sec x + tan x. x+tan x R √x e dx √ Try the substitution w = x 1 dw = 2√1 x dx = 2w dx, dx = 2w dw R √x R w R e dx = e 2w dw = 2 we w dw Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I ln x dx I Integration by parts ; let u = ln x, dv = dx ...... R I tan x dx R R sin x I write as dx, let u = cos x...... tan x dx = cos x R I sin3 x cos x dx I Substitution, let u = sin x...... R I √ 1 dx 2 25−x I I I I I I I I “ ” R −1 x √ 1 Trig. substitution, x = 5 sin θ, − π2 ≤ θ ≤ π2 , dx = sin 5 25−x 2 R sec x dx R R x(sec x+tan x) sec xdx = sec sec Let u = sec x + tan x. x+tan x R √x e dx √ Try the substitution w = x 1 dw = 2√1 x dx = 2w dx, dx = 2w dw R √x R w R e dx = e 2w dw = 2 we w dw Use integration by parts now with u = w , dv = e w dw . Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I sin(7x) cos(4x) dx Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I sin(7x) cos(4x) dx ˆ ˜ I Use sin(mx) cos(nx) = 1 sin((m − n)x) + sin((m + n)x 2 Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I sin(7x) cos(4x) dx ˆ ˜ I Use sin(mx) cos(nx) = 1 sin((m − n)x) + sin((m + n)x 2 R I cos2 x dx Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I sin(7x) cos(4x) dx ˆ ˜ I Use sin(mx) cos(nx) = 1 sin((m − n)x) + sin((m + n)x 2 R I cos2 x dx I Use the half angle formula: cos2 x = 21 (1 + cos 2x). Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I sin(7x) cos(4x) dx ˆ ˜ I Use sin(mx) cos(nx) = 1 sin((m − n)x) + sin((m + n)x 2 R I cos2 x dx I I Use the half angle formula: cos2 x = 21 (1 + cos 2x). R 1 dx x 2 −9 Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I sin(7x) cos(4x) dx ˆ ˜ I Use sin(mx) cos(nx) = 1 sin((m − n)x) + sin((m + n)x 2 R I cos2 x dx I I I Use the half angle formula: cos2 x = 21 (1 + cos 2x). R 1 dx x 2 −9 R A B Partial Fractions x 21−9 dx = x−3 + x+3 Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I sin(7x) cos(4x) dx ˆ ˜ I Use sin(mx) cos(nx) = 1 sin((m − n)x) + sin((m + n)x 2 R I cos2 x dx I I I I Use the half angle formula: cos2 x = 21 (1 + cos 2x). R 1 dx x 2 −9 R A B Partial Fractions x 21−9 dx = x−3 + x+3 R x dx x+3 Annette Pilkington Strategy for Integration Standard Table Review Outline How you would approach the following integrals: R I sin(7x) cos(4x) dx ˆ ˜ I Use sin(mx) cos(nx) = 1 sin((m − n)x) + sin((m + n)x 2 R I cos2 x dx I I I I I Use the half angle formula: cos2 x = 21 (1 + cos 2x). R 1 dx x 2 −9 R A B Partial Fractions x 21−9 dx = x−3 + x+3 R x dx x+3 R x R dx = u−3 du where u = x + 3. x+3 u Annette Pilkington Strategy for Integration Standard Table More Challenging Integrals The following integrals may require multiple steps: R x2 I dx 9+x 6 Annette Pilkington Strategy for Integration Standard Table More Challenging Integrals The following integrals may require multiple steps: R x2 I dx 9+x 6 I Substitute u = x 3 , du = 3x 2 dx Annette Pilkington Strategy for Integration Standard Table More Challenging Integrals The following integrals may require multiple steps: R x2 I dx 9+x 6 I I Substitute u = x 3 , du = 3x 2 dx R x2 R 1 dx = 31 9+u 2 du 9+x 6 Annette Pilkington Strategy for Integration Standard Table More Challenging Integrals The following integrals may require multiple steps: R x2 I dx 9+x 6 I Substitute u = x 3 , du = 3x 2 dx R x2 R 1 dx = 31 9+u 2 du 9+x 6 I Use tan−1 formula or substitute u = tan θ. I Annette Pilkington Strategy for Integration Standard Table More Challenging Integrals The following integrals may require multiple steps: R x2 I dx 9+x 6 I I I I Substitute u = x 3 , du = 3x 2 dx R x2 R 1 dx = 31 9+u 2 du 9+x 6 Use tan−1 formula or substitute u = tan θ. R 1 dx x 2 +27x+26 Annette Pilkington Strategy for Integration Standard Table More Challenging Integrals The following integrals may require multiple steps: R x2 I dx 9+x 6 I I Substitute u = x 3 , du = 3x 2 dx R x2 R 1 dx = 31 9+u 2 du 9+x 6 I Use tan−1 formula or substitute u = tan θ. R 1 dx x 2 +27x+26 I Partial Fractions x 2 + 27x + 26 = (x + 26)(x + 1). I Annette Pilkington Strategy for Integration Standard Table More Challenging Integrals The following integrals may require multiple steps: R x2 I dx 9+x 6 I I I I I I Substitute u = x 3 , du = 3x 2 dx R x2 R 1 dx = 31 9+u 2 du 9+x 6 Use tan−1 formula or substitute u = tan θ. R 1 dx x 2 +27x+26 Partial Fractions x 2 + 27x + 26 = (x + 26)(x + 1). R x arctan x dx (1+x 2 )2 Annette Pilkington Strategy for Integration Standard Table More Challenging Integrals The following integrals may require multiple steps: R x2 I dx 9+x 6 I I I I Substitute u = x 3 , du = 3x 2 dx R x2 R 1 dx = 31 9+u 2 du 9+x 6 Use tan−1 formula or substitute u = tan θ. R 1 dx x 2 +27x+26 I Partial Fractions x 2 + 27x + 26 = (x + 26)(x + 1). R x arctan x dx (1+x 2 )2 I integration by parts with u = arctanx and dv = I Annette Pilkington x . (1+x 2 )2 Strategy for Integration Standard Table More Challenging Integrals The following integrals may require multiple steps: R x2 I dx 9+x 6 I I I I Substitute u = x 3 , du = 3x 2 dx R x2 R 1 dx = 31 9+u 2 du 9+x 6 Use tan−1 formula or substitute u = tan θ. R 1 dx x 2 +27x+26 I Partial Fractions x 2 + 27x + 26 = (x + 26)(x + 1). R x arctan x dx (1+x 2 )2 I integration by parts with u = arctanx and dv = I 1 and du = 1+x R 2x R v = (1+x 2 )2 dx = (w = 1 + x 2 ) = I Annette Pilkington 1 2 · 1 w2 = x . (1+x 2 )2 −1 2w = Strategy for Integration −1 2(1+x 2 ) . Standard Table More Challenging Integrals The following integrals may require multiple steps: R x2 I dx 9+x 6 I I I I Substitute u = x 3 , du = 3x 2 dx R x2 R 1 dx = 31 9+u 2 du 9+x 6 Use tan−1 formula or substitute u = tan θ. R 1 dx x 2 +27x+26 I Partial Fractions x 2 + 27x + 26 = (x + 26)(x + 1). R x arctan x dx (1+x 2 )2 I integration by parts with u = arctanx and dv = I 1 and du = 1+x R 2x R v = (1+x 2 )2 dx = (w = 1 + x 2 ) = 12 · R x arctan x R 1 dx = −(arctanx) + 2(1+x 2 )2 dx (1+x 2 )2 2(1+x 2 ) I I Annette Pilkington 1 w2 = x . (1+x 2 )2 −1 2w = Strategy for Integration −1 2(1+x 2 ) . Standard Table More Challenging Integrals The following integrals may require multiple steps: R x2 I dx 9+x 6 I I I I Substitute u = x 3 , du = 3x 2 dx R x2 R 1 dx = 31 9+u 2 du 9+x 6 Use tan−1 formula or substitute u = tan θ. R 1 dx x 2 +27x+26 I Partial Fractions x 2 + 27x + 26 = (x + 26)(x + 1). R x arctan x dx (1+x 2 )2 I integration by parts with u = arctanx and dv = I 1 and du = 1+x R 2x R v = (1+x 2 )2 dx = (w = 1 + x 2 ) = 12 · R x arctan x R 1 dx = −(arctanx) + 2(1+x 2 )2 dx (1+x 2 )2 2(1+x 2 ) I I I 1 w2 = x . (1+x 2 )2 −1 2w = −1 2(1+x 2 ) . latter integral use trig substitution x = tan θ, we get RFor the R sec −2 x dx = cos2 x dx, we can use the half angle formula. Annette Pilkington Strategy for Integration Standard Table More More Challenging Integrals The following integrals may require multiple steps: R I √ ln x dx 2 x 1+(ln x) Annette Pilkington Strategy for Integration Standard Table More More Challenging Integrals The following integrals may require multiple steps: R I √ ln x dx 2 x I 1+(ln x) u = ln x followed by w = 1 + u 2 Annette Pilkington Strategy for Integration Standard Table More More Challenging Integrals The following integrals may require multiple steps: R I √ ln x dx 2 x 1+(ln x) I u = ln x followed by w = 1 + u 2 I R 1+sin x dx. 1−sin x (Requires True Grit Annette Pilkington ) Strategy for Integration Standard Table More More Challenging Integrals The following integrals may require multiple steps: R I √ ln x dx 2 x 1+(ln x) I u = ln x followed by w = 1 + u 2 I R I 1+sin x dx. 1−sin x multiply by (Requires True Grit ) 1+sin x . 1+sin x Annette Pilkington Strategy for Integration Standard Table More More Challenging Integrals The following integrals may require multiple steps: R I √ ln x dx 2 x 1+(ln x) I u = ln x followed by w = 1 + u 2 I R I I 1+sin x dx. 1−sin x (Requires True Grit ) 1+sin x . 1+sin x multiply by R 1+sin x 1+sin x R · 1+sin x dx = 1−sin x (1+sin x)2 1−sin2 x Annette Pilkington dx = R (1+sin x)2 cos2 x dx. Strategy for Integration Standard Table More More Challenging Integrals The following integrals may require multiple steps: R I √ ln x dx 2 x 1+(ln x) I u = ln x followed by w = 1 + u 2 I R I I I 1+sin x dx. 1−sin x (Requires True Grit ) 1+sin x . 1+sin x multiply by R 1+sin x 1+sin x R R x)2 x)2 · 1+sin x dx = (1+sin dx = (1+sin dx. 1−sin x 1−sin2 x cos2 x R (1+sin x)2 R 1+2 sin x+sin2 x R 1 R R dx = dx = cos2 x dx + 2cossin2 xx dx + cos2 x cos2 x Annette Pilkington Strategy for Integration sin2 x cos2 x dx. Standard Table More More Challenging Integrals The following integrals may require multiple steps: R I √ ln x dx 2 x 1+(ln x) I u = ln x followed by w = 1 + u 2 I R I I I I 1+sin x dx. 1−sin x (Requires True Grit ) 1+sin x . 1+sin x multiply by R 1+sin x 1+sin x R R x)2 x)2 · 1+sin x dx = (1+sin dx = (1+sin dx. 1−sin x 1−sin2 x cos2 x R (1+sin x)2 R 1+2 sin x+sin2 x R 1 R R dx = dx = cos2 x dx + 2cossin2 xx dx + 2 2 R cos x2 R 2 sin xcos x R = sec x dx + cos2 x dx + tan2 x dx = R R tan x − u22 du + sec2 x − 1 dx, where u = cos x. Annette Pilkington Strategy for Integration sin2 x cos2 x dx. Standard Table More More Challenging Integrals The following integrals may require multiple steps: R I √ ln x dx 2 x 1+(ln x) I u = ln x followed by w = 1 + u 2 I R I I I I I 1+sin x dx. 1−sin x (Requires True Grit ) 1+sin x . 1+sin x multiply by R 1+sin x 1+sin x R R x)2 x)2 · 1+sin x dx = (1+sin dx = (1+sin dx. 1−sin x 1−sin2 x cos2 x R (1+sin x)2 R 1+2 sin x+sin2 x R 1 R R dx = dx = cos2 x dx + 2cossin2 xx dx + 2 2 R cos x2 R 2 sin xcos x R = sec x dx + cos2 x dx + tan2 x dx = R R tan x − u22 du + sec2 x − 1 dx, where u = cos x. = 2 tan x + 2 sec x − x + C Annette Pilkington Strategy for Integration sin2 x cos2 x dx. Standard Table More More Challenging Integrals The following integrals may require multiple steps: R I √ ln x dx 2 x 1+(ln x) I u = ln x followed by w = 1 + u 2 I R I I I I I I 1+sin x dx. 1−sin x (Requires True Grit ) 1+sin x . 1+sin x multiply by R 1+sin x 1+sin x R R x)2 x)2 · 1+sin x dx = (1+sin dx = (1+sin dx. 1−sin x 1−sin2 x cos2 x R (1+sin x)2 R 1+2 sin x+sin2 x R 1 R R sin2 x dx = dx = cos2 x dx + 2cossin2 xx dx + cos 2 2 2 x dx. R cos x2 R 2 sin xcos x R 2 = sec x dx + cos2 x dx + tan x dx = R R tan x − u22 du + sec2 x − 1 dx, where u = cos x. = 2 tan x + 2 sec x − x + C Note if you integrate this in Mathematica you get a different looking answer, but both differ by a constant In[10]:= Out[10]= x x SimplifyB - x CosB F + H4 + xL SinB F ì 2 2 x x CosB F - SinB F 2 2 - H2 Tan@xD + 2 Sec@xD - xLF -2 Annette Pilkington Strategy for Integration Standard Table Even More More Challenging Integrals The following integrals may require multiple steps: R ln x I √ dx x Annette Pilkington Strategy for Integration Standard Table Even More More Challenging Integrals The following integrals may require multiple steps: R ln x I √ dx x √ 1 1 I Let u = x, du = √ dx = 2u dx 2 x Annette Pilkington Strategy for Integration Standard Table Even More More Challenging Integrals The following integrals may require multiple steps: R ln x I √ dx x √ 1 1 I Let u = x, du = √ dx = 2u dx 2 x R ln x R 2 I Let √ dx = 2 ln u du x Annette Pilkington Strategy for Integration Standard Table Even More More Challenging Integrals The following integrals may require multiple steps: R ln x I √ dx x √ 1 1 I Let u = x, du = √ dx = 2u dx 2 x R ln x R 2 I Let √ dx = 2 ln u du x R R I = 4 ln u du, we use integration by parts on ln u du. Annette Pilkington Strategy for Integration