Solutions to Integration Using a Power Substitution

SOLUTIONS TO INTEGRATION USING A POWER SUBSTITUTION

SOLUTION 11 : Integrate $\displaystyle{ \int { 1 \over \sqrt{ 2 + \sqrt{ 1 + \sqrt{x} } } } \, dx }$ . Remove the ``outside" square root first. Use the power substitution

$2 + \sqrt{ 1 + \sqrt{x} } = u^2$

so that

$u = \sqrt{ 2 + \sqrt{ 1 + \sqrt{x} } }$

$\longrightarrow \sqrt{ 1 + \sqrt{x} } = u^2-2$

$\longrightarrow \sqrt{x} = (u^2-2)^2-1$

$\longrightarrow x = ((u^2-2)^2-1)^2$ ,

and (Use the chain rule.)

Substitute into the original problem, replacing all forms of , getting

$\displaystyle{ \int { 1 \over \sqrt{ 2 + \sqrt{ 1 + \sqrt{x} } } } \, dx } = \displaystyle{ \int { 1 \over u } \, (8u^7-48u^5+88u^3-48u) du }$

$= \displaystyle{ \int {1 \over u}u(8u^6-48u^4+88u^2-48) \, du }$

$= \displaystyle{ \int (8u^6-48u^4+88u^2-48) \, du }$

$= \displaystyle{ 8{u^7 \over 7} - 48{u^5 \over 5} + 88{u^3 \over 3} - 48u } + C$

$= \displaystyle{ {8 \over 7}\bigg(\sqrt{ 2 + \sqrt{ 1 + \sqrt{x} } }\bigg)^7 -... ...+ \sqrt{ 1 + \sqrt{x} } }\bigg)^3 - 48\sqrt{ 2 + \sqrt{ 1 + \sqrt{x} } } } + C$

$= \displaystyle{ {8 \over 7}\bigg(\bigg( 2 + \sqrt{ 1 + \sqrt{x} } \bigg)^{1/2... ...+ \sqrt{x} } \bigg)^{1/2}\bigg)^3 - 48\sqrt{ 2 + \sqrt{ 1 + \sqrt{x} } } } + C$

$= \displaystyle{ {8 \over 7}\bigg( 2 + \sqrt{ 1 + \sqrt{x} } \bigg)^{7/2} - {4... ...qrt{ 1 + \sqrt{x} } \bigg)^{3/2} - 48 \sqrt{ 2 + \sqrt{ 1 + \sqrt{x} } } } + C$ .

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SOLUTION 12 : Integrate $\displaystyle{ \int { \sqrt{x} \over x-1 } \, dx }$ . Use the power substitution

so that

$\sqrt{x} = \sqrt{u^2} = u$

and

Substitute into the original problem, replacing all forms of , getting

$\displaystyle{ \int { \sqrt{x} \over x-1 } \, dx } = \displaystyle{ \int { u \over u^2-1 } \, (2u) du }$

$= \displaystyle{ \int { 2u^2 \over u^2-1 } \, du }$

(Use polynomial division.)

$= \displaystyle{ \int \Big( 2 + {2 \over u^2-1} \Big) \, du }$ .

Use the method of partial fractions. Factor and decompose into partial fractions, getting

$\displaystyle{ \int \Big( 2 + {2 \over u^2-1} \Big) \, du } = \displaystyle{ \int \Big( 2 + {2 \over (u-1)(u+1) } \Big) \, du }$

$= \displaystyle{ \int { \Big( 2 + {A \over u-1} + {B \over u+1}\Big)} \,dx}$

(After getting a common denominator, adding fractions, and equating numerators, it follows that $\ \ A(u+1) + B(u-1) = 2$ ;

let $\displaystyle{u = -1: \ A(0) + B(-2) = 2 \longrightarrow B = -1 }$ ;

let $\displaystyle{u = 1: \ A(2) + B(0) = 2 \longrightarrow A = 1 }$ .)

$= \displaystyle{ \int { \Big( 2 + {1 \over u-1} - {1 \over u+1} \Big)} \,dx}$

$= \displaystyle{ 2u + \ln \vert u-1\vert - \ln \vert u+1\vert + C}$

$= \displaystyle{ 2\sqrt{x} + \ln \vert\sqrt{x}-1\vert - \ln \vert\sqrt{x}+1\vert + C}$

(Recall that $\ \ln m - \ln n = \ln \Big( \displaystyle{ m \over n } \Big)$ .)

$= \displaystyle{ 2\sqrt{x} + \ln { \vert\sqrt{x}-1\vert \over \vert\sqrt{x}+1\vert } + C }$ .

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SOLUTION 13 : Integrate $\displaystyle{ \int { \sqrt{4-x} \over x^2 } \, dx }$ . Use the power substitution

so that

$\sqrt{4-x} = u$ ,

and

Substitute into the original problem, replacing all forms of , getting

$\displaystyle{ \int { \sqrt{4-x} \over x^2 } \, dx } = \displaystyle{ \int { u \over (4-u^2)^2 } \, (-2u) du }$

$= \displaystyle{ \int { -2u^2 \over (u^2-4)^2 } \, du }$ .

Use the method of partial fractions. Factor and decompose into partial fractions, getting (There are repeated linear factors!)

$\displaystyle{ \int { -2u^2 \over (u^2-4)^2 } \, du } = \displaystyle{ \int { -2u^2 \over ((u-2)(u+2))^2 } \, du }$

$= \displaystyle{ \int { -2u^2 \over (u-2)^2(u+2)^2 } \, du }$

$= \displaystyle{ \int { \Big( {A \over u-2} + {B \over (u-2)^2} + {C \over u+2} + {D \over (u+2)^2 } \Big)} \,du }$

(After getting a common denominator, adding fractions, and equating numerators, it follows that

;

let $\displaystyle{u = 2: \ A(0) + B(16) + C(0) + D(0) = -8 \longrightarrow B = -{1 \over 2}}$ ;

let $\displaystyle{u = -2: \ A(0) + B(0) + C(0) + D(16) = -8 \longrightarrow D = -{1 \over 2}}$ ;

let $\displaystyle{u = 0: \ A(-8) + B(4) + C(8) + D(4) = -8A - 2 + 8C - 2 = 0 }$

$\ \ \ \ \ \ \ \ \displaystyle{\longrightarrow A - C = -{1 \over 2} }$ ;

let $\displaystyle{u = 1: \ A(-9) + B(9) + C(3) + D(1) = -9A - {9 \over 2} + 3C - {1 \over 2} = -2 }$

$\ \ \ \ \ \ \ \ \displaystyle{ \longrightarrow -3A + C = 1 }$ ;

it follows that $\ A = -\displaystyle{1 \over 4}$ and $C = \displaystyle{1 \over 4}$ .)

$= \displaystyle{ \int { \Big( {-1/4 \over u-2} + {-1/2 \over (u-2)^2} + {1/4 \over u+2} + {-1/2 \over (u+2)^2 } \Big)} \,du }$

$= \displaystyle{ \int { \Big( -(1/4){1 \over u-2} - (1/2)(u-2)^{-2} + (1/4){1 \over u+2} - (1/2)(u+2)^{-2} \Big)} \,du }$

$= \displaystyle{ -{1 \over 4} \ln \vert u-2\vert - {1 \over 2}{(u-2)^{-1} \ove... ... } + {1 \over 4}\ln \vert u+2\vert - {1 \over 2}{(u+2)^{-1} \over (-1) } } + C$

$= \displaystyle{ {1 \over 4}\ln \vert u+2\vert - {1 \over 4} \ln \vert u-2\vert + {1 \over 2(u-2)} + {1 \over 2(u+2) } } + C$

$= \displaystyle{ {1 \over 4} \Big( \ln \vert u+2\vert - \ln \vert u-2\vert \Big) + {1 \over 2(u-2)} + {1 \over 2(u+2) } } + C$

(Recall that $\ \ln m - \ln n = \ln \Big( \displaystyle{ m \over n } \Big)$ .)

$= \displaystyle{ {1 \over 4} \ln { \vert u+2\vert \over \vert u-2\vert } + {1 \over 2(u-2)} + {1 \over 2(u+2) } } + C$

$= \displaystyle{ {1 \over 4} \ln { \vert\sqrt{4-x}+2\vert \over \vert\sqrt{4-x}-2\vert } + {1 \over 2(\sqrt{4-x}-2)} + {1 \over 2(\sqrt{4-x}+2) } } + C$ .

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SOLUTION 14 : Integrate $\displaystyle{ \int { x^{1/4} + 5 \over x - 16 } \, dx }$ . Use the power substitution

so that

$x^{1/4} = u$

and

Substitute into the original problem, replacing all forms of , getting

$\displaystyle{ \int { x^{1/4} + 5 \over x - 16 } \, dx } = \displaystyle{ \int { u + 5 \over u^4 - 16 } \, (4u^3) du }$

$= \displaystyle{ \int { 4u^4 + 20u^3 \over u^4 - 16 } \, du }$

(Use polynomial division.)

$= \displaystyle{ \int { \Big( 4 + {20u^3 + 64 \over u^4 - 16} \Big) } \, du }$ .

Use the method of partial fractions. Factor and decompose into partial fractions, getting

$\displaystyle{ \int { \Big( 4 + {20u^3 + 64 \over u^4 - 16} \Big) } \, du } = ... ...style{ \int { \Big( 4 + {20u^3 + 64 \over (u^2 - 4)(u^2 + 4) } \Big) } \, du }$

$= \displaystyle{ \int { \Big( 4 + {20u^3 + 64 \over (u-2)(u+2)(u^2 + 4) } \Big) } \, du }$

$= \displaystyle{ \int { \Big( 4 + {A \over u-2} + {B \over u+2} + {Cu+D \over u^2+4 } \Big) } \,du }$

(After getting a common denominator, adding fractions, and equating numerators, it follows that

;

let $\displaystyle{u = -2: \ A(0) + B(-32) + (C(-2)+D)(0) = -96 \longrightarrow B = 3}$ ;

let $\displaystyle{u = 2: \ A(32) + B(0) + (C(2)+D)(0) = 224 \longrightarrow A = 7 }$ ;

let $\displaystyle{u = 2i: \ A(2i+2)(0) + B(2i-2)(0) + (C(2i)+D)(2i+2)(2i-2) }$

$\ \ \ \ \ \ \ \ = (2Ci+D)(-4-4)= -16Ci-8D = 20(2i)^3+64$

$\displaystyle{\longrightarrow -16Ci-8D = 20(8i^3)+64 = 160(-i)+64 = -160i+64 }$ ;

it follows that $-16C = -160 \ \$ and $\ \ -8D = 64 \ \longrightarrow C = 10$ and .)

$= \displaystyle{ \int { \Big( 4 + {7 \over u-2} + {3 \over u+2} + {10u-8 \over u^2+4 } \Big)} \,du }$

$= \displaystyle{ \int { \Big( 4 + 7{1 \over u-2} + 3{1 \over u+2} + {10u \over u^2+4 } + {-8 \over u^2+4 } \Big)} \,du }$

$= \displaystyle{ \int { \Big( 4 + 7{1 \over u-2} + 3{1 \over u+2} + 10{u \over u^2+4 } - 8{1 \over u^2+2^2 } \Big)} \,du }$

$= \displaystyle{ 4u + 7 \ln\vert u-2\vert + 3 \ln\vert u+2\vert + 10\Big({1 \o... ...2}\Big) \ln\vert u^2+4\vert - 8 \Big({1 \over 2}\Big) \arctan{u \over 2} } + C$

$= \displaystyle{ 4x^{1/4} + 7 \ln\vert x^{1/4}-2\vert + 3 \ln\vert x^{1/4}+2\vert + 5 \ln\vert(x^{1/4})^2+4\vert - 4 \arctan {x^{1/4} \over 2} } + C$

$= \displaystyle{ 4x^{1/4} + 7 \ln\vert x^{1/4}-2\vert + 3 \ln\vert x^{1/4}+2\vert + 5 \ln\vert x^{1/2}+4\vert - 4 \arctan {x^{1/4} \over 2} } + C$ .

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About this document ...

Duane Kouba 2000-05-09