How do I find the indefinite integral for $\int \frac{6}{2x-x^2}dx$?

Question

How can I integrate this $\int\frac{6}{2x-x^2}dx$ ?

I know I have to integrate it using logarithms, I just don't know how to do this one. Can someone help me out?

Answer 1

HINT:

Apply partial fraction decomposition method. $$\int\frac{6}{2x-x^2}dx=3\int\left[\frac{1}{x}+\frac{1}{(2-x)}\right]dx$$

Answer 2

$$\int \frac{6}{2x-x^2}\, dx =6\int\frac{d(x-1)}{1-(x-1)^2}=6\ \frac12\ln\left|\frac{1+(x-1)}{1-(x-1)}\right|+C=3\ln\left|\frac{x}{2-x}\right| +C$$

Answer 3

Complete the square:

$$2x-x^2= 1-(x^2-2x+1)=1-(x-1)^2,$$

then substitute $u=x-1 \implies du=dx$ so that the integral becomes

$$6\int \frac{1}{1-u^2}\,du.$$

Can you finish?