<s>[INST] Integral of $\frac{\sqrt {x}}{x^2+x}$ I'm trying to find $$\int_{1/3}^3 \frac{\sqrt{x}}{x^2+x} dx.$$
I used a $u$ substitution where $u = \sqrt{x}$ to get $$2 \int_\sqrt{1/3}^\sqrt{3} \frac{u}{u^4+u^2} du.$$
Substituting $u = \tan v$, I get
$$2 \int \cot v dv
  = \left. \ln |\sin v|\right|_{\arctan \sqrt{1/3}}^{\arctan \sqrt 3}
  = 2\ln \left( \frac{\sqrt{3}}{2}\right)-2\ln(1/2),
$$ which is wrong. Where did I go wrong?
 [/INST] Your mistake is in your first substitution. You replaced $dx$ with $2 du$ instead of $2u\,du$. The substitution, by the way, makes the integral trivial:
\begin{align}
\int_{1/3}^3 \frac{\sqrt{x}}{x^2+x} dx&=\int_{1/\sqrt 3}^{\sqrt3}\frac{2u^2}{u^4+u^2}\,du
=\int_{1/\sqrt 3}^{\sqrt3}\frac{2}{1+u^2}\,du\\ \ \\ 
&=2\left.\vphantom\int\arctan u\,\right|_{1/\sqrt3}^{\sqrt3}
=2\arctan\sqrt3-2\arctan1/\sqrt3\\ \ \\
&=\frac{2\pi}3-\frac{2\pi}6=\frac\pi3.
\end{align}
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