## Problem 1

Calculate: $\int \frac{3 x^{4}+2 x^{3}+x^{2}-2015}{\left(x^{4}+x^{3}+x^{2}+2015\right)^{2}} d x, x \in \mathbb{R}$.
## Problem 1

We are looking for $a, b \in \mathbb{R}$ such that $\left(\frac{a x+b}{x^{4}+x^{3}+x^{2}+2015}\right)^{\prime}=\frac{3 x^{4}+2 x^{3}+x^{2}-2015}{\left(x^{4}+x^{3}+x^{2}+2015\right)^{2}}$

$\Leftrightarrow \frac{-3 a x^{4}+(-2 a-4 b) x^{3}+(-a-3 b) x^{2}+(-2 b) x+2015 a}{\left(x^{4}+x^{3}+x^{2}+2015\right)^{2}}=\frac{3 x^{4}+2 x^{3}+x^{2}-2015}{\left(x^{4}+x^{3}+x^{2}+2015\right)^{2}}$

Identifying the coefficients, we get $a=-1, b=0 \Rightarrow\left(\frac{-x}{x^{4}+x^{3}+x^{2}+2015}\right)^{\prime}=\frac{3 x^{4}+2 x^{3}+x^{2}-2015}{\left(x^{4}+x^{3}+x^{2}+2015\right)^{2}} \ldots \ldots . .1 \mathbf{p}$

$\Rightarrow \int \frac{3 x^{4}+2 x^{3}+x^{2}-2015}{\left(x^{4}+x^{3}+x^{2}+2015\right)^{2}} d x=\int\left(\frac{-x}{x^{4}+x^{3}+x^{2}+2015}\right)^{\prime} d x=\frac{-x}{x^{4}+x^{3}+x^{2}+2015}+C$