Final answer to the problem
$\frac{x^{4}}{4}+\frac{-x^{3}}{3}-\frac{1}{2}x^2+2x-4\arctan\left(x\right)+\frac{1}{2}\ln\left(x^2+1\right)+C_0$
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Step-by-step Solution
Specify the solving method
1
Find the integral
$\int\frac{x^5-x^4+x^2-2}{x^2+1}dx$
2
Divide $x^5-x^4+x^2-2$ by $x^2+1$
$\begin{array}{l}\phantom{\phantom{;}x^{2}+1;}{\phantom{;}x^{3}-x^{2}-x\phantom{;}+2\phantom{;}\phantom{;}}\\\phantom{;}x^{2}+1\overline{\smash{)}\phantom{;}x^{5}-x^{4}\phantom{-;x^n}+x^{2}\phantom{-;x^n}-2\phantom{;}\phantom{;}}\\\phantom{\phantom{;}x^{2}+1;}\underline{-x^{5}\phantom{-;x^n}-x^{3}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{-x^{5}-x^{3};}-x^{4}-x^{3}+x^{2}\phantom{-;x^n}-2\phantom{;}\phantom{;}\\\phantom{\phantom{;}x^{2}+1-;x^n;}\underline{\phantom{;}x^{4}\phantom{-;x^n}+x^{2}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{;\phantom{;}x^{4}+x^{2}-;x^n;}-x^{3}+2x^{2}\phantom{-;x^n}-2\phantom{;}\phantom{;}\\\phantom{\phantom{;}x^{2}+1-;x^n-;x^n;}\underline{\phantom{;}x^{3}\phantom{-;x^n}+x\phantom{;}\phantom{-;x^n}}\\\phantom{;;\phantom{;}x^{3}+x\phantom{;}-;x^n-;x^n;}\phantom{;}2x^{2}+x\phantom{;}-2\phantom{;}\phantom{;}\\\phantom{\phantom{;}x^{2}+1-;x^n-;x^n-;x^n;}\underline{-2x^{2}\phantom{-;x^n}-2\phantom{;}\phantom{;}}\\\phantom{;;;-2x^{2}-2\phantom{;}\phantom{;}-;x^n-;x^n-;x^n;}\phantom{;}x\phantom{;}-4\phantom{;}\phantom{;}\\\end{array}$
3
Resulting polynomial
$\int\left(x^{3}-x^{2}-x+2+\frac{x-4}{x^2+1}\right)dx$
4
Expand the integral $\int\left(x^{3}-x^{2}-x+2+\frac{x-4}{x^2+1}\right)dx$ into $5$ integrals using the sum rule for integrals, to then solve each integral separately
$\int x^{3}dx+\int-x^{2}dx+\int-xdx+\int2dx+\int\frac{x-4}{x^2+1}dx$
5
The integral $\int x^{3}dx$ results in: $\frac{x^{4}}{4}$
$\frac{x^{4}}{4}$
6
The integral $\int-x^{2}dx$ results in: $\frac{-x^{3}}{3}$
$\frac{-x^{3}}{3}$
7
The integral $\int-xdx$ results in: $-\frac{1}{2}x^2$
$-\frac{1}{2}x^2$
8
The integral $\int2dx$ results in: $2x$
$2x$
9
The integral $\int\frac{x-4}{x^2+1}dx$ results in: $\frac{1}{2}\ln\left(x^2+1\right)-4\arctan\left(x\right)$
$\frac{1}{2}\ln\left(x^2+1\right)-4\arctan\left(x\right)$
10
Gather the results of all integrals
$\frac{x^{4}}{4}+\frac{-x^{3}}{3}-\frac{1}{2}x^2+2x-4\arctan\left(x\right)+\frac{1}{2}\ln\left(x^2+1\right)$
11
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\frac{x^{4}}{4}+\frac{-x^{3}}{3}-\frac{1}{2}x^2+2x-4\arctan\left(x\right)+\frac{1}{2}\ln\left(x^2+1\right)+C_0$
Final answer to the problem
$\frac{x^{4}}{4}+\frac{-x^{3}}{3}-\frac{1}{2}x^2+2x-4\arctan\left(x\right)+\frac{1}{2}\ln\left(x^2+1\right)+C_0$