How to integrate $\int \frac{2x^2+x}{(x+1)(x^2+1)}dx$? How to integrate $\displaystyle \int \frac{2x^2+x}{(x+1)(x^2+1)}dx$? I Tried using partial fractions but i got lost, thanks.

Partial fractions are the way to go. The fraction is already reduced, and the denominator is fully factored over the reals, so your setup is
$$\frac{2x^2+x}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}=\frac{A(x^2+1)+(Bx+C)(x+1)}{(x+1)(x^2+1)}\;,$$
and you must find $A,B$, and $C$ so that $$2x^2+x=A(x^2+1)+(Bx+C)(x+1)=(A+B)x^2+(B+C)x+(A+C)\;.$$
Equating coefficients of powers of $x$ yields the system
$$\left\{\begin{align*}
&A+B=2\\
&B+C=1\\
&A+C=0\;,
\end{align*}\right.$$
which is easily solved: $A=\frac12,B=\frac32$, and $C=-\frac12$. Thus,
$$\frac{2x^2+x}{(x+1)(x^2+1)}=\frac1{2(x+1)}+\frac{3x-1}{2(x^2+1)}\;,$$ and
$$\int\frac{2x^2+x}{(x+1)(x^2+1)}dx=\frac12\int\frac1{x+1}dx+\frac12\int\frac{3x-1}{x^2+1}dx\;.$$
You shouldn’t have any trouble with $\int\frac1{x+1}dx$. The other term is most easily handled by splitting it:
$$\int\frac{3x-1}{x^2+1}dx=3\int\frac{x}{x^2+1}dx-\int\frac1{x^2+1}dx\;,$$
where the first integral succumbs to a $u$-substitution, and the second is one that you should know (or at least be able to work by a trig substitution).
