How do you find the indefinite integral of int (5x^2 – 4x + 7)/((x^2+1)(2x – 1)) dx?

1 Answer
Oct 18, 2015

I=-2arctanx+5/2ln|2x-1| +C

Explanation:

I=int (5x^2 – 4x + 7)/((x^2+1)(2x – 1)) dx

(Ax+B)/(x^2+1)+C/(2x – 1)=((Ax+B)(2x-1)+C(x^2+1))/((x^2+1)(2x – 1)) =

=(2Ax^2+2Bx-Ax-B+Cx^2+C)/((x^2+1)(2x – 1)) =

=(x^2(2A+C)+x(2B-A)+(-B+C))/((x^2+1)(2x – 1)) = T

If we compare integrand with expression T, we can write:

2A+C=5
2B-A=-4 => A=2B+4
-B+C=7 => C=B+7

2(2B+4)+B+7=5
5B=-10 => B=-2
A=0
C=5

I=int (-2/(x^2+1)+5/(2x-1))dx

I=-2arctanx+5/2ln|2x-1| +C