Download 4x dx 2x 3 + 1 dw w ln w 5 ln(5) ln(3) ln 3 = = ) = − = then dw 4xdx = =

MATH 141 – Practice Gateway Exam #1 Solutions
The use of calculators or any computer algebra system is not permitted on this test. No partial
credit will be awarded. A score of 7 is required to pass this test.
Compute each of the following.
1)
3
∫ (2x + x + 12)dx
x 4 x2
+ + 12x + C
4 2
1
1
= x 4 + x2 + 12x + C
2
2
=2
2)
1
Let w = 2x2 + 3 then dw = 4x dx
⌠
4x

dx

2
⌡ 2x + 3
x = 0 → w = 2(0)2 + 3 = 3
0
x = 1 → w = 2(1)2 + 3 = 5
5
⌠
1
dw
⌡w
= 
3
= [ln w ]
5
3
5
= ln(5) − ln(3) = ln  
3
3)
∫ 3e
( 5x )dx
Let w = 5x then dw = 5dx
w 1
= ∫ 3e( ) dw
5
3
w
= ∫ e( ) dw
5
3
= ew + C
5
3
= e5x + C
5
4)
⌠


⌡
4
4
A
B
dx
= +
x(x − 2)
x(x − 2) x x − 2
⌠  −2
2 
=   +
dx
x − 2 
⌡ x
= −2 ln x − ln x − 2  + C
4 = A(x − 2) + B(x)
4 = (A + B)x − 2A
4 = −2A
0 = A + B = −2 + B
so A = −2
so B = 2
5)
⌠
 3x −2

⌡
1
+  dx
x
x −1
=3
+ ln x + C
−1
= −3x −1 + ln x + C
6)
∫ 8x cos(x)dx
= 8 [ x ⋅ sin(x) − ∫ sin(x)dx ]
Let
u=x
du = dx
dv = cos(x)dx
v = sin(x)
= 8 [ x sin(x) + cos(x)] + C
7)
3 4
6
∫ 2x (x + 5) dx
Let w = x 4 + 5 then
⌠
1
= 2 (w)6 ⋅ dw
⌡
4
dw = 4x3 dx
1
dw = x 3 dx
4
1 w7
= 2⋅
+C
4 7
(x 4 + 5)7
=
+C
14
8)
Let w = 4x + 7 then
∫ 4x + 7 dx
1
⌠
=  (4x + 7)2
⌡
dx
1
1⌠
=  (w)2 dw
4⌡
3
2
w
1
= ⋅
+C
4 3
2
3
1 2
= ⋅ (4x + 7)2 + C
4 3
3
1
= (4x + 7)2 + C
6
dw = 4dx
1
dw = dx
4