If $\int \frac{1}{x^3+x^4}dx=\frac{A}{x^2}+\frac{B}{x}+\log \frac{x}{|x+1|}+C,$ then $2A+2B=$

Solution:

We have,

$\begin{array}{l} I=\int \frac{1}{x^3+x^4}dx=\int \frac{1}{x^3(x+1)}dx=\int \frac{1}{x^2\left(x\right(x+1\left)\right)}dx\\ \Rightarrow I=\int \frac{1}{x^2}\left(\frac{1}{x}-\frac{1}{x+1}\right)dx=\int \frac{1}{x}\left\{\frac{1}{x^2}-\frac{1}{x(x+1)}\right\}dx\\ \Rightarrow I=\int \frac{1}{x}\left\{\frac{1}{x^2}-\left(\frac{1}{x}-\frac{1}{x+1}\right)\right\}dx=\int \frac{1}{x^3}-\frac{1}{x^2}+\frac{1}{x(x+1)}dx\\ \Rightarrow I=\int \left(\frac{1}{x^3}-\frac{1}{x^2}+\frac{1}{x}-\frac{1}{x+1}\right)dx=-\frac{1}{2}\frac{1}{x^2}+\frac{1}{x}+\log \left|\frac{x}{x+1}\right|+C\end{array}$

$\therefore A=-\frac{1}{2}\text{ and }B=1\Rightarrow 2A+2B=1$