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SOLUTIONS TO INTEGRATION USING A POWER SUBSTITUTION



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

$ x = u^2 $

so that

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

and

$ dx = (2u) du $ .

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

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

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

(Use polynomial division.)

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

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

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

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

Click HERE to return to the list of problems.




SOLUTION 2 : Integrate $ \displaystyle{ \int { 2+\sqrt{x} \over 3-\sqrt{x} } \,dx } $ . Use the power substitution

$ x = u^2 $

so that

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

and

$ dx = (2u) du $ .

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

$ \displaystyle{ \int { 2+\sqrt{x} \over 3-\sqrt{x} } \,dx } = \displaystyle{ \int { 2+u \over 3-u } \, (2u)du } $

$ = \displaystyle{ \int { 2u^2+4u \over 3-u } \, du } $

(Use polynomial division.)

$ = \displaystyle{ \int \Big( -2u-10 + {30 \over 3-u} \Big) \, du } $

$ = \displaystyle{ \int \Big( -2u-10 + 30{1 \over 3-u} \Big) \, du } $

$ = \displaystyle{ -u^2-10u + 30(-1) \ln \vert 3-u\vert } + C $

$ = \displaystyle{ -x-10\sqrt{x} - 30 \ln \vert 3-\sqrt{x}\vert } + C $ .

Click HERE to return to the list of problems.




SOLUTION 3 : Integrate $ \displaystyle{ \int { 3 \over 4+x^{1/3} } \,dx } $ . Use the power substitution

$ x = u^3 $

so that

$ x^{1/3} = (u^3)^{1/3} = u $

and

$ dx = (3u^2) du $ .

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

$ \displaystyle{ \int { 3 \over 4+x^{1/3} } \,dx } = \displaystyle{ \int { 3 \over 4+u } \, (3u^2) du } $

$ = \displaystyle{ \int { 9u^2 \over u+4 } \, du } $

(Use polynomial division.)

$ = \displaystyle{ \int \Big( 9u-36 + {144 \over u+4} \Big) \, du } $

$ = \displaystyle{ \int \Big( 9u-36 + 144{1 \over u+4} \Big) \, du } $

$ = \displaystyle{ 9{u^2 \over 2} -36u + 144 \ln \vert u+4\vert } + C $

$ = \displaystyle{ {9 \over 2}(x^{1/3})^2 - 36x^{1/3} + 144 \ln \vert x^{1/3}+4\vert } + C $

$ = \displaystyle{ {9 \over 2}x^{2/3} - 36x^{1/3} + 144 \ln \vert x^{1/3}+4\vert } + C $ .

Click HERE to return to the list of problems.




SOLUTION 4 : Integrate $ \displaystyle{ \int { x(x-1)^{1/6} } \, dx } $ . Use the power substitution

$ x-1 = u^6 $

so that

$ x = u^6 + 1 $ ,

$ (x-1)^{1/6} = (u^6)^{1/6} = u $ ,

and

$ dx = (6u^5) du $ .

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

$ \displaystyle{ \int { x(x-1)^{1/6} } \, dx } = \displaystyle{ \int { (u^6 + 1)u } \, (6u^5) du } $

$ = \displaystyle{ \int (6u^{12} + 6u^6) \, du } $

$ = \displaystyle{ 6{u^{13} \over 13} + 6{u^{7} \over 7} } + C $

$ = \displaystyle{ {6 \over 13}((x-1)^{1/6})^{13} + {6 \over 7}((x-1)^{1/6})^{7} } + C $

$ = \displaystyle{ {6 \over 13}(x-1)^{13/6} + {6 \over 7}(x-1)^{7/6} } + C $ .

Click HERE to return to the list of problems.




SOLUTION 5 : Integrate $ \displaystyle{ \int { 3x+2 \over \sqrt{x-9} } \, dx } $ . Use the power substitution

$ x-9 = u^2 $

so that

$ x = u^2 + 9 $ ,

$ \sqrt{x-9} = \sqrt{u^2} = u $ ,

and

$ dx = (2u) du $ .

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

$ \displaystyle{ \int { 3x+2 \over \sqrt{x-9} } \, dx } = \displaystyle{ \int { 3(u^2 + 9)+2 \over u } \, (2u) du } $

$ = \displaystyle{ \int (6u^{2} + 58) \, du } $

$ = \displaystyle{ 6{u^{3} \over 3} + 58u } + C $

$ = \displaystyle{ 2(\sqrt{x-9})^{3} + 58\sqrt{x-9} } + C $

$ = \displaystyle{ 2((x-9)^{1/2})^{3} + 58\sqrt{x-9} } + C $

$ = \displaystyle{ 2(x-9)^{3/2} + 58\sqrt{x-9} } + C $ .

Click HERE to return to the list of problems.




SOLUTION 6 : Integrate $ \displaystyle{ \int { 1 \over x^{2/3} - x^{1/3} } \, dx } $ . Use the power substitution

$ x = u^3 $

so that

$ x^{1/3} = (u^3)^{1/3} = u $ ,

$ x^{2/3} = (u^3)^{2/3} = u^2 $ ,

and

$ dx = (3u^2) du $ .

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

$ \displaystyle{ \int { 1 \over x^{2/3} - x^{1/3} } \, dx } = \displaystyle{ \int { 1 \over u^2 - u } \, (3u^2) du } $

$ = \displaystyle{ \int { 3u^2 \over u(u-1) } \, du } $

$ = \displaystyle{ \int { 3u \over u-1 } \, du } $

(Use polynomial division.)

$ = \displaystyle{ \int \Big( 3 + {3 \over u-1} \Big) \, du } $

$ = \displaystyle{ \int \Big( 3 + 3{1 \over u-1} \Big) \, du } $

$ = \displaystyle{ 3u + 3 \ln \vert u-1\vert } + C $

$ = \displaystyle{ 3x^{1/3} + 3 \ln \vert x^{1/3}-1\vert } + C $ .

Click HERE to return to the list of problems.





Duane Kouba 2000-05-09