﻿ U-substitution and the General Power Rule Worksheet | Problems & Solutions

# U-substitution and the General Power Rule Worksheet

U-substitution and the General Power Rule Worksheet
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1.
Evaluate ${\int }_{}^{}$($x$ + 9)$d$$x$.
 a. ($\frac{2x}{5}$ + $\frac{26}{3}$) + C b. ($\frac{x}{5}$ + $\frac{26}{3}$) $\frac{{\left(x-10\right)}^{\frac{3}{2}}}{3}$ + C c. ($\frac{2x}{5}$ + $\frac{26}{3}$) $\frac{{\left(x-10\right)}^{\frac{3}{2}}}{3}$ + C d. ($\frac{2x}{5}$ + $\frac{26}{3}$) $\frac{{\left(x-10\right)}^{\frac{5}{2}}}{3}$ + C

#### Solution:

(x + 9)x - 10dx = (u2 + 19)u (2u)du
[Let u² = x - 10, 2u du = dx.]

= (2u4 + 38u2) du

= 2(u55) + 38(u33) + C

= 2((x-10)525) + 38((x-10)323) + C
[Substitute u2 = x - 10.]

= (x-10)323[25(x-10) + 383]

= [2x5 + 26 / 3] (x-10)323 + C

2.
Evaluate ${\int }_{}^{}$$\frac{{x}^{3}}{{\left(7+{x}^{2}\right)}^{3}}$$d$$x$.
 a. ($\frac{7}{4}$) $\frac{1}{{\left(7+{x}^{2}\right)}^{2}}$ + C b. $\frac{-1}{2\left(7+{x}^{2}\right)}$ + ($\frac{7}{4}$) $\frac{1}{{\left(7+{x}^{2}\right)}^{2}}$ + C c. $\frac{-1}{2\left(7+{x}^{2}\right)}$ + C d. does not exist

#### Solution:

x3(7+x2)3dx = x2(7+x2)3x dx
[Let u2 = 7 + x2, u du = x dx.]

= (u-3 - 7u-5)du

= (- 1 / 2)u-2 + 7 / 4u- 4 + C

= -12(7+x2) + (7 / 4) 1(7+x2)2 + C
[Substitute u2 = 7 + x2.]

3.
Evaluate ${\int }_{}^{}$ 16$x$.
 a. $\frac{8}{7}$ (${e}^{{14x}^{2}}$) + C b. $\frac{4}{7}$ (${e}^{{14x}^{2}}$) + C c. 8 ${x}^{2}{e}^{{14x}^{2}}$ + C d. (${e}^{{14x}^{2}}$) + C

#### Solution:

16xe14x2 dx = 16 / 28 xe14x2 1x du
[Let u = 14x2 du = 28xdx.]

= 4 / 7 eu du
[Substitute 14x2 = u.]

= 4 / 7 eu + C

= 4 / 7 (e14x2) + C
[Substitute u = 14x2.]

4.
Evaluate ${\int }_{}^{}$.
 a. 46 ln |7 - ${x}^{2}$| + C b. 23 ln |7 - ${x}^{2}$| + C c. $\frac{1}{2}$ ln |7 - ${x}^{2}$| + C d. ln |7 - ${x}^{2}$| + C

#### Solution:

- 46x7 -x2 dx

Let u = 7 - x2 du = - 2x dx

= 23 1u du

= 23 ln |u| + c

= 23 ln |7 - x2| + C
[Substitute u = 7 - x2.]

5.
${\int }_{}^{}$$d$$x$ =
 a. - + C b. - $\frac{155}{{\left({x}^{3}+8\right)}^{6}}$ + C c. - + C d. - + C

#### Solution:

Let x3 + 8 = u 3x2dx = du.

So, 31x2(x3 + 8)5 dx = 31x2(x3 + 8)5 13x2 du
[As du = 3x2 dx dx = 13x2 du.]

= 31 / 3 1u5 du
[u = x3 + 8.]

= 31 / 3 u- 5du

= 31 / 3 (u- 4- 4) + C

= - 31 / 12(x3 + 8)- 4 + C
[Substitute u = x3 + 8.]

= - 3112(x3 + 8)4 + C

6.
Evaluate ${\int }_{}^{}$ $d$$x$.
 a. + C b. + C c. 10 + C d. $\frac{1}{22}$ + C

#### Solution:

Let u = x2 - 32x du = (2x - 32) dx

So, (x2 - 32x)10(2x - 32)dx = (x2 - 32x)10(2x - 32) 1(2x - 32) du
[As du = (2x - 32) dx dx = du2x - 32 .]

= u10 du
[u = x2 - 32x.]

= u1111 + C

= (x2 - 32x)1111 + C
[Substitute u = x2 - 32x.]

7.
Evaluate ${\int }_{}^{}$.
 a. 19 ln |tan $x$ - 1| + C b. ln |tan ($\frac{x}{2}$) + 1| + C c. 19 ln |tan ($\frac{x}{2}$) - 1| + C d. ln |tan $x$ - 1| + C

#### Solution:

Let u = tan(x2) x = 2 tan-1 (u) and dx = 21+u2du

sin x = 2u1+u2, cos x = 1-u21+u2

So, 19sin x - cos x - 1 dx = 192u1+u2-1-u21+u2 - 121+u2 du
[Substitute u = tanx2 and dx = 21+u2 du.]

= 19u - 1 du

= 19ln |u -1| + C

= 19ln |tan(x2 ) - 1| + C
[Substitute u = tan(x2 ).]

8.
${\int }_{}^{}$ =
 a. + C b. + C c. 40coth($\frac{x}{2}$) + C d. + C

#### Solution:

Let u = tanh (x / 2) x = 2 tanh-1u and dx = 21 -u2 du

Sinh x = 2u1 -u2, cosh x = 1 +u21 -u2

So, 20sinh x + cosh x dx = 202u1 -u2 +1 +u21 -u221 -u2 du

= 40(1 + u)2 du

= - 40u + 1 + C

= - 40tanh(x2) + 1 + C
[Substitute u = tanh(x / 2).]

9.
Find ${\int }_{}^{}$$d$$x$.
 a. ln |4$x$2 + 7$x$ + 19| + C b. 41 log |4$x$2 + 7$x$ + 19| + C c. 41 ln |4$x$2 + 7$x$ + 19| + C d. $\frac{-41}{{\left(4{x}^{2}+7x+19\right)}^{2}}$ + C

#### Solution:

Let (4x2 + 7x + 19) = u (8x + 7) dx = du

So, 41(8x + 7)4x2 + 7x + 19 dx = 41 8x + 74x2 + 7x + 19 (18x + 7) du
[As (8x + 7) dx = du dx = du8x + 7.]

= 411u du
[Substitute 4x2 + 7x + 19 = u.]

= 41 ln |u| + c

= 41 ln |4x2 + 7x + 19| + C
[Substitute u = 4x2 + 7x + 19.]

10.
Evaluate ${\int }_{}^{}$ 4$x$ .
 a. + C b. $\frac{2}{7}$ + C c. 4 ${e}^{7{x}^{2}+27}$ + C d. $\frac{1}{14}$ + C

#### Solution:

Let 7x2 + 27 = u 14x dx = du

So, 4x e7x2 + 27 dx = 4 / 14 (xe7x2 + 27)1x du
[As 14x dx = du dx = 114x du.]

= 2 / 7 eu du
[Substitute 7x2 + 27 = u]

= 2 / 7 eu + C

= 2 / 7 e7x2 + 27 + C
[Substitute u = 7x2 + 27.]