How do you find $int (x^2+2x)/((x^2+1)^2)dx$ using partial fractions?

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Answer 1 · 2017-02-02T07:04:09

$int(2x-1)/(x^2+1)^2dx=1/2(tan^(-1)x)+(x-2)/(x^2+1)+c$

Partial fractions of $(x^2+2x)/(x^2+1)^2$ , will be of type

$(x^2+2x)/(x^2+1)^2=(Ax+B)/(x^2+1)+(Cx+D)/(x^2+1)^2$

= $((Ax+B)(x^2+1)+Cx+D)/(x^2+1)^2$

= $(Ax^3+Bx^2+Ax+B+Cx+D)/(x^2+1)^2$

= $(Ax^3+Bx^2+(A+C)x+(B+D))/(x^2+1)^2$

Comparing the terms, we get $A=0$ , $B=1$ , $A+C=2$ and $B+D=0$

i.e. $A=0$ , $B=1$ , $C=2$ and $D=-1$

Hence, $(x^2+2x)/(x^2+1)^2=1/(x^2+1)+(2x-1)/(x^2+1)^2$ and

$int(x^2+2x)/(x^2+1)^2dx=int1/(x^2+1)dx+int(2x-1)/(x^2+1)^2dx$

Now $int1/(x^2+1)dx=tan^(-1)x$ is a well known integral and for the other one

We can break $int(2x-1)/(x^2+1)^2dx=int(2x)/(x^2+1)^2dx-int1/(x^2+1)^2dx$

For first term, let us assume $u=x^2+1$ , then $du=2xdx$ = $int(2x)/(x^2+1)^2dx=int(du)/u^2=-1/u=-1/(x^2+1)$

and for second term let us assume $x=tanv$ then

$int1/(x^2+1)^2dx=int(sec^2v)/sec^4vdv=intcos^2vdv$

= $1/2int(1+cos2v)dv=v/2+1/4sin2v$

= $tan^(-1)x/2+1/2sinvcosv$

= $tan^(-1)x/2+1/2x/sqrt(x^2+1)xx1/sqrt(x^2+1)$

= $tan^(-1)x/2+1/2x/(x^2+1)$

Hence, $int(2x-1)/(x^2+1)^2dx$

= $-1/(x^2+1)+tan^(-1)x/2+1/2x/(x^2+1)+c$

= $1/2(tan^(-1)x)+(x-2)/(x^2+1)+c$

How do you find int (x^2+2x)/((x^2+1)^2)dxint (x^2+2x)/((x^2+1)^2)dx using partial fractions?