Answer to Question #124472 in Calculus for TUHIN SUBHRA DAS

Question

integrate x^2+x+5/(x^2+4)(x+1)dx

Answer 1

Let 5/((x²+4)(x+1))=(Ax+B)/(x²+4)+C/(x+1).

Then 5=(Ax+B)(x+1)+C(x²+4)

5=(A+C)x²+(B+A)x+(B+4C)

We obtain system

A+C=0,

B+A=0,

B+4C=5.

Then B=-A, C=-A=>-A-4A=5=>-5A=-5=>A=-1,B=1,C=1.

And 5/((x²+4)(x+1))=(-x+1)/(x²+4)+1/(x+1).

So ∫ (x²+x+5/((x²+4)(x+1)))dx=$\int$ (x²+x+(-x+1)/(x²+4)+1/(x+1))dx=

$\int$(x²+x-x/(x²+4)+1/(x²+4)+1/(x+1))dx=x³/3+x²/2-(ln(x²+4))/2+(arctan(x/2))/2+ln(x+1)+C