Help with Calculate integral

Question

Find $\int^a_0 \dfrac{3x^2-ax}{(x-2a)(x^2+a^2)} dx$ I tried using partial fractions and the substitution $u=a-x$ but I haven't made any real progress. Please help.

Answer 1

Use partial fraction $$\frac{3x^2-ax}{(x-2a)(x^2+a^2)}=\frac{A}{(x-2a)}+\frac{Bx+C}{(x^2+a^2)}$$ Now find A,B,C $$ A=2\\B=1\\C=a\\\frac{3x^2-ax}{(x-2a)(x^2+a^2)}=\frac{2}{(x-2a)}+\frac{x+a}{(x^2+a^2)}=\\\frac{3x^2-ax}{(x-2a)(x^2+a^2)}=\frac{2}{(x-2a)}+\frac{x}{(x^2+a^2)}+\frac{a}{(x^2+a^2)}$$ So $$\int_0^a \frac{3x^2-ax}{(x-2a)(x^2+a^2)}dx=\\ \int_0^a (\frac{2}{(x-2a)}+\frac{x}{(x^2+a^2)}+\frac{a}{(x^2+a^2)})dx = \\ \left(2\ln|x-2a| +\frac{1}{2} \ln(x^2+a^2) + \tan^{-1}(\frac{x}{a})\right) \Big|_0^a.$$

Answer 2

Hint: Use the formular:$$\frac{3x^2-ax}{(x-2a)(x^2+a^2)}=\frac{A}{(x-2a)}+\frac{Bx+C}{(x^2+a^2)}$$ to resolve $\frac{3x^2-ax}{(x-2a)(x^2+a^2)}$ into partial fraction by finding the values of A, B and C. Hope you can do that?

Then $$\int_0^a \frac{3x^2-ax}{(x-2a)(x^2+a^2)}dx = \int_0^a \frac{A}{(x-2a)}dx +\int_0^a \frac{Bx+C}{(x^2+a^2)}dx.$$