Integrate $ \int \frac{4x^2+2x}{(3x-1)(x^2+1)}dx$.

Question

$$ \int \frac{4x^2+2x}{(3x-1)(x^2+1)}dx$$

I am having difficulty determining which integration technique to use for the above question. I have tried partial fraction decomposition with two different approaches but fail to get the correct answer. I also attempted integration by parts but was unable to get to a place where I could integrate.

Does anyone have a suggestion as to which integration technique would work best for solving this problem?

Thank you!

Answer 1

Hint: partial fraction decomposition yields $$ \frac{4x^{2}+2x}{(3x-1)(x^{2}+1)}=\frac{x+1}{x^{2}+1}+\frac{1}{3x-1}. $$ You can check your answer using the following: $$ \int\frac{x+1}{x^{2}+1}=\frac{1}{2}\log(x^{2}+1)+\tan^{-1}(x)+C_{1}\text{ and }\int\frac{1}{3x-1}=\frac{1}{3}\log(3x-1)+C_{2}. $$

Answer 2

Noting that the partial fractions method is indeed the way to go, we set up the problem first by trying to find the partial fractions:

$\dfrac{4x^2+2x}{(3x-1)(x^2+1)}=\dfrac{A}{3x-1}+\dfrac{Bx+C}{x^2+1}$

I will let you do the intermediary steps, but the partial fractions comes out to be $A=1; B=1; C=1$

$\int \frac{x+1}{x^2+1}dx$ seems tricky but divide this integral into $\int\frac{x}{x^2+1}dx+\int\frac{1}{x^2+1}dx$