007B Sample Midterm 2, Problem 5 Detailed Solution

Evaluate the integral:

$\int {\frac {4x}{(x+1)(x^{2}+1)}}~dx$

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Background Information:
Through partial fraction decomposition, we can write
${\frac {1}{(x+1)(x^{2}+1)}}={\frac {A}{x+1}}+{\frac {Bx+C}{x^{2}+1}}$
for some constants $A,B.$

Solution:

Step 1:
We need to use partial fraction decomposition for this integral.
To start, we let
${\frac {4x}{(x+1)(x^{2}+1)}}={\frac {A}{x+1}}+{\frac {Bx+C}{x^{2}+1}}.$
Multiplying both sides of the last equation by  $(x+1)(x^{2}+1),$
we get
$4x=A(x^{2}+1)+(Bx+C)(x+1).$

Step 2:
If we let  $x=-1,$  the last equation becomes  $-4=2A.$  So,  $A=-2.$
If we let  $x=0,$  then we get  $0=A+C.$  Thus,  $C=-A=2.$
Finally, if we let  $x=1,$  we get  $4=2A+2B+2C.$
Plugging in  $A=-2$  and  $C=2,$  we get  $B=2.$
So, in summation, we have
${\frac {4x}{(x+1)(x^{2}+1)}}={\frac {-2}{x+1}}+{\frac {2x+2}{x^{2}+1}}.$

Step 3:
Now, we have
${\begin{array}{rcl}\displaystyle {\int {\frac {4x}{(x+1)(x^{2}+1)}}~dx}&=&\displaystyle {\int {\frac {-2}{x+1}}~dx+\int {\frac {2x+2}{x^{2}+1}}~dx}\\&&\\&=&\displaystyle {\int {\frac {-2}{x+1}}~dx+\int {\frac {2x}{x^{2}+1}}~dx+\int {\frac {2}{x^{2}+1}}~dx}\\&&\\&=&\displaystyle {\int {\frac {-2}{x+1}}~dx+\int {\frac {2x}{x^{2}+1}}~dx+2\arctan(x).}\end{array}}$
For the remaining integrals, we use  $u$-substitution.
For the first integral, we substitute  $u=x+1.$
For the second integral, the substitution is  $t=x^{2}+1.$
Then, we integrate to get
${\begin{array}{rcl}\displaystyle {\int {\frac {4x}{(x+1)(x^{2}+1)}}~dx}&=&\displaystyle {\int {\frac {-2}{u}}~du+\int {\frac {1}{t}}~dt+2\arctan(x)}\\&&\\&=&\displaystyle {-2\ln |u|+\ln |t|+2\arctan(x)+C}\\&&\\&=&\displaystyle {-2\ln |x+1|+\ln |x^{2}+1|+2\arctan(x)+C.}\end{array}}$

Final Answer:
$-2\ln |x+1|+\ln |x^{2}+1|+2\arctan(x)+C$

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