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SOLUTION 9 : Integrate $ \displaystyle{ \int { 1 \over x^2+4x+5 } \,dx } $ . First, complete the square in the denominator. The result is

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

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

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

Now use u-substitution. Let

$ u = x+2 $

so that

$ du = (1) dx = dx $ .

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

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

(Use formula 2 from the introduction to this section.)

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

$ = \displaystyle{ \arctan (x+2) } + C $ .

Click HERE to return to the list of problems.




SOLUTION 10 : Integrate $ \displaystyle{ \int { 1 \over 2x^2+12x+68 } \,dx } $ . First, factor 2 from the denominator. The result is

$ \displaystyle{ \int { 1 \over 2x^2+12x+68 } \,dx }
= \displaystyle{ \int { 1 \over 2(x^2+6x+34) } \,dx } $

$ = \displaystyle{ (1/2) \int { 1 \over x^2+6x+34 } \,dx } $

(Complete the square in the denominator.)

$ = \displaystyle{ (1/2) \int { 1 \over (x^2+6x)+34 } \,dx } $

$ = \displaystyle{ (1/2) \int { 1 \over (x^2+6x+9)-9+34 } \,dx } $

$ = \displaystyle{ (1/2) \int { 1 \over (x+3)^2+25 } \,dx } $

$ = \displaystyle{ (1/2) \int { 1 \over (x+3)^2+5^2 } \,dx } $ .

Use u-substitution. Let

$ u = x+3 $

so that

$ du = (1) dx = dx $ .

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

$ \displaystyle{ (1/2) \int { 1 \over (x+3)^2+5^2 } \,dx }
= \displaystyle{ (1/2) \int { 1 \over u^2+5^2 } \,du } $

(Use formula 3 from the introduction to this section.)

$ = \displaystyle{ (1/2) (1/5) \arctan (u/5) } + C $

$ = \displaystyle{ {1 \over 10} \arctan \Big( {x+3 \over 10} \Big) } + C $ .

Click HERE to return to the list of problems.




SOLUTION 11 : Integrate $ \displaystyle{ \int { 1 \over \sqrt{x} (x+9) } \,dx } $ . Because of the term $ \sqrt{x} $ in the denominator, rewrite the term $ x+9 $ in a somewhat unusual way. The result is

$ \displaystyle{ \int { 1 \over \sqrt{x} (x+9) } \,dx }
= \displaystyle{ \int { 1 \over \sqrt{x} \big( \big( \sqrt{x} \big)^2+9 \big) } \,dx } $ .

Now use u-substitution. Let

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

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 $ .

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

$ \displaystyle{ \int { 1 \over \sqrt{x} \big( \big( \sqrt{x} \big)^2+9 \big) } ...
...playstyle{ \int { 1 \over \big( \sqrt{x} \big)^2+9 } \,{1 \over \sqrt{x}} dx } $

$ = \displaystyle{ \int { 1 \over u^2+9 } \,(2) du } $

$ = \displaystyle{ 2 \int { 1 \over u^2+3^2 } \, du } $

(Use formula 3 from the introduction to this section.)

$ = \displaystyle{ 2 (1/3) \arctan (u/3) } + C $

$ = \displaystyle{ {2 \over 3} \arctan \Big( { \sqrt{x} \over 3 } \Big) } + C $ .

Click HERE to return to the list of problems.




SOLUTION 12 : Integrate $ \displaystyle{ \int { \sin (2x) \over 1 + \cos (2x) } \,dx } $ . Use u-substitution. Let

$ u = 1 + \cos (2x) $

so that (Don't forget to use the chain rule on $ \cos (2x) $.)

$ du = -\sin (2x) (2) dx = -2 \sin (2x) dx $ ,

or

$ (-1/2) du = \sin (2x) dx $ .

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

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

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

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

(Use formula 1 from the introduction to this section.)

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

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

Click HERE to return to the list of problems.




SOLUTION 13 : Integrate $ \displaystyle{ \int { \cos x \over 1 + \sin^2 x } \,dx } $ . First, rewrite the denominator of the function, getting

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

Now use u-substitution. Let

$ u = \sin x $

so that

$ du = \cos x \ dx $ .

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

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

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

(Use formula 2 from the introduction to this section.)

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

$ = \displaystyle{ \arctan ( \sin x) } + C $ .

Click HERE to return to the list of problems.




SOLUTION 14 : Integrate $ \displaystyle{ \int { e^{2x} \over 1 + e^{2x} } \,dx } $ . Use u-substitution. Let

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

so that (Don't forget to use the chain rule on $ e^{2x} $.)

$ du = e^{2x} (2) dx = 2 e^{2x} dx $ ,

or

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

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

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

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

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

(Use formula 1 from the introduction to this section.)

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

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

Click HERE to return to the list of problems.




SOLUTION 15 : Integrate $ \displaystyle{ \int { e^x \over 1 + e^{2x} } \,dx } $ . First, rewrite the denominator of the function, getting (Recall that $ A^{mn} = (A^m )^n $ .)

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

Now use u-substitution. Let

$ u = e^{x} $

so that

$ du = e^{x} dx $ .

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

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

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

(Use formula 2 from the introduction to this section.)

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

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

Click HERE to return to the list of problems.




SOLUTION 16 : Integrate $ \displaystyle{ \int {e^{4x} \over 1 + e^{2x} } \,dx } $ . Use u-substitution. Let

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

so that

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

or

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

In addition, we can "back substitute" with

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

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

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

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

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

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

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

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

$ = (1/2) \big( u - \ln \vert u\vert \big) + C $

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

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

(Combine $ 1/2 $ and $ C $ since $ C $ is an arbitrary constant.)

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

Click HERE to return to the list of problems.




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Duane Kouba 2000-04-12