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SOLUTION 17 : Integrate $ \displaystyle{ \int { x \over x^2 + 2x + 1 } \,dx } $ . First factor the denominator, getting

$ \displaystyle{ \int { x \over x^2 + 2x + 1 } \,dx } =\displaystyle{ \int { x \over (x + 1)^2 } \,dx } $ .

Now use u-substitution. Let

$ u = x+1 $

so that

$ du = (1) dx = dx $ .

In addition, we can "back substitute" with

$ x = u-1 $ .

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

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

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

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

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

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

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

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SOLUTION 18 : Integrate $ \displaystyle{ \int { x \over x^2 + 2x + 5 } \,dx } $ . First complete the square in the denominator, getting

$ \displaystyle{ \int { x \over x^2 + 2x + 5 } \,dx } = \displaystyle{ \int { x \over (x^2 + 2x + 1) +4 } \,dx } $

$ = \displaystyle{ \int { x \over (x + 1)^2 +4 } \,dx } $ .

Now use u-substitution. Let

$ u = x+1 $

so that

$ du = (1) dx = dx $ .

In addition, we can "back substitute" with

$ x = u-1 $ .

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

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

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

$ = \displaystyle{ \int { u \over u^2 + 4} \,du - \int { 1 \over u^2 + 4} \,du } $ .

In the first integral use substitution. Let

$ w = u^2 + 4 $

so that

$ dw = 2u du $ ,

or

$ (1/2) dw = u du $ .

Substitute into the first integral, replacing all forms of $ u $, and use formula 3 from the beginning of this section on the second integral, getting

$ \displaystyle{ \int { u \over u^2 + 4} \,du - \int { 1 \over u^2 + 4} \,du } ... ...isplaystyle{ \int { 1 \over u^2 + 4} \,u du - \int { 1 \over u^2 + 2^2} \,du } $

$ = \displaystyle{ \int { 1 \over w} \, (1/2) dw} - (1/2) \arctan (u/2) + C $

$ = \displaystyle{ (1/2) \int { 1 \over w} \, dw} - (1/2) \arctan ((x+1)/2) + C $

$ = \displaystyle{ (1/2) \ln \vert w\vert - (1/2) \arctan ((x+1)/2) + C } $

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

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

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SOLUTION 19 : Integrate $ \displaystyle{ \int { 7x - 2\over 2x^2 - 16x + 42 } \,dx } $ . First factor out a 2 and complete the square in the denominator, getting

$ \displaystyle{ \int { 7x - 2\over 2x^2 - 16x + 42 } \,dx } = \displaystyle{ \int { 7x - 2\over 2(x^2 - 8x + 21) } \,dx } $

$ = (1/2)\displaystyle{ \int { 7x - 2\over (x^2 - 8x) + 21 } \,dx }$

$ = (1/2)\displaystyle{ \int { 7x - 2\over (x^2 - 8x + 16) - 16 + 21 } \,dx }$

$ = (1/2)\displaystyle{ \int { 7x - 2\over (x-4)^2 + 5 } \,dx }$ .

Now use u-substitution. Let

$ u = x-4 $

so that

$ du = (1) dx = dx $ .

In addition, we can "back substitute" with

$ x = u+4 $ .

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

$ (1/2)\displaystyle{ \int { 7x - 2\over (x - 4)^2 +5 } \,dx } = (1/2)\displaystyle{ \int { 7(u+4) - 2 \over u^2 + 5} \, du } $

$ = (1/2)\displaystyle{ \int { 7u + 26 \over u^2 + 5} \, du } $

$ = (1/2)\displaystyle{ \int {\Big\{ {7u \over u^2 + 5 } + { 26 \over u^2 + 5 } \Big\} } \, du} $

$ = \displaystyle{ (7/2) \int { u \over u^2 + 5} \,du + (13)\int { 1 \over u^2 + 5} \,du } $ .

In the first integral use substitution. Let

$ w = u^2 + 5 $

so that

$ dw = 2u du $ ,

or

$ (1/2) dw = u du $ .

Substitute into the first integral, replacing all forms of $ u $, and use formula 3 from the beginning of this section on the second integral, getting

$ \displaystyle{ (7/2) \int { u \over u^2 + 5} \,du + (13)\int { 1 \over u^2 + 5... ...{ 1 \over u^2 + 5} u \,du + (13)\int { 1 \over u^2 + ( \sqrt{5} )^2 } \,du } $

$ = \displaystyle{ (7/2) \int { 1 \over w} \, (1/2) dw + (13)(1/ \sqrt{5} ) \arctan (u/ \sqrt {5} ) + C } $

$ = \displaystyle{ (7/2)(1/2) \int { 1 \over w} \, dw + (13/ \sqrt{5} ) \arctan ((x-4)/ \sqrt {5} ) + C } $

$ = \displaystyle{ (7/4)\ln \vert w\vert + (13/ \sqrt{5} ) \arctan ((x-4)/ \sqrt {5} ) + C } $

$ = \displaystyle{ (7/4)\ln \vert u^2 + 5\vert + (13/ \sqrt{5} ) \arctan ((x-4)/ \sqrt {5} ) + C } $

$ = \displaystyle{ (7/4)\ln \vert(x-4)^2 + 5\vert + (13/ \sqrt{5} ) \arctan ((x-4)/ \sqrt {5} )+ C } $ .

Click HERE to return to the list of problems.




SOLUTION 20 : Integrate $ \displaystyle{ \int { 1 \over 1 + e^{2x} } \,dx } $ . First rewrite this rational function by multiplying by $ \displaystyle{ e^{-2x} \over e^{-2x} } $, getting

$ \displaystyle{ \int { 1 \over 1 + e^{2x} } \,dx } = \displaystyle{ \int { 1 \over 1 + e^{2x} } { e^{-2x} \over e^{-2x} } \,dx } $

$ = \displaystyle{ \int { e^{-2x} \over e^{-2x} + e^{2x}e^{-2x} } \,dx } $

(Recall that $ A^m A^n = A^{m+n} $ .)

$ = \displaystyle{ \int { e^{-2x} \over e^{-2x} + e^{2x-2x} } \,dx } $

$ = \displaystyle{ \int { e^{-2x} \over e^{-2x} + e^{0} } \,dx } $

$ = \displaystyle{ \int { e^{-2x} \over e^{-2x} + 1 } \,dx } $ .

Now use substitution. Let

$ u = e^{-2x} + 1 $

so that

$ du = -2e^{-2x} dx $ ,

or

$ (-1/2) du = e^{-2x} dx $ .

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

$ \displaystyle{ \int { e^{-2x} \over e^{-2x} + 1 } \,dx } = \displaystyle{ \int { 1 \over e^{-2x} + 1 } e^{-2x} \,dx } $

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

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

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

$ = \displaystyle{ (-1/2) \ln\vert e^{-2x} + 1\vert + C } $ .

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SOLUTION 21 : Integrate $ \displaystyle{ \int { e^{3x} \over 1 + e^{2x} } \,dx } $ . Use u-substitution. Let

$ u = e^{x} $

so that

$ du = e^{x} dx $ .

Now rewrite this rational function using rules of exponents. Then

$ \displaystyle{ \int { e^{3x} \over 1 + e^{2x} } \,dx = \int { e^{2x+x} \over 1 + (e^{x})^2} \,dx} $

$ = \displaystyle{ \int { e^{2x} e^{x} \over 1 + (e^{x})^2 } \,dx } $

$ = \displaystyle{ \int { (e^{x})^2 e^{x} \over 1 + (e^{x})^2 } \,dx } $

$ = \displaystyle{ \int { (e^{x})^2 \over 1 + (e^{x})^2 } e^{x} \,dx } $ .

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

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

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

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

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

$ = \displaystyle{ u - \arctan u + C } $

$ = \displaystyle{ e^x - \arctan (e^x) + C } $ .

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SOLUTION 22 : Integrate $ \displaystyle{ \int { 9 + 6 \sqrt {x} + x \over 4 \sqrt {x} +x } \,dx } $ . First rewrite this rational function as

$ \displaystyle{ \int { 9 + 6 \sqrt x + x \over 4 \sqrt x +x } \,dx = \int { (\sqrt{x} + 3)^2 \over \sqrt{x} (4 + \sqrt{x})} \,dx} $ .

Now use u-substitution. Let

$ u = 4 + \sqrt{x} $.

so that

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

or

$ 2 du = \displaystyle{ 1 \over \sqrt{x} } dx $ .

In addition, we can "back substitute" with

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

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

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

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

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

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

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

$ = 2\displaystyle{ \Big\{ {u^2 \over 2} - 2u + \ln \vert u\vert \Big\} + C} $

$ = \displaystyle{ {u^2 } - 4u + 2\ln \vert u\vert + C} $

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

$ = 16 + 8\sqrt {x} + x - 16 -4 \sqrt {x} + 2\ln \vert 4 + \sqrt{x}\vert + C $

$ = x + 4 \sqrt {x} + 2\ln \vert 4 + \sqrt{x}\vert + C $ .

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Duane Kouba 2000-04-12