Computing the Integral $\int \frac{\sqrt{x}}{x^2+x} dx$ 
Find $\displaystyle \int \dfrac{\sqrt{x}}{x^2+x} dx$.

What would be the best way to integrate this? I saw the answer to this and it looked simple so that might mean the steps would be too?
 A: Hint
Use the substitution
$$x=u^2$$
A: Here is how I did it: put $x = t^2$ and your integral turns into
$$\int \dfrac t {t^4 + t^2} 2t \Bbb d t = 2 \int \frac 1 {1 + t^2} \Bbb d t = 2 \arctan t + c = 2 \arctan \sqrt x + c .$$
Hope it Helps!
A: Say $x=z^2$ and $dx=2zdz$
Hence the integrand stands $$\int \frac{z}{z^4+z^2} \cdot 2z dz$$
$$= 2\int \frac{z^2}{z^2(z^2+1)} dz$$
$$= 2\int \frac{dz}{z^2+1} $$
$$= 2 \arctan z +c$$
$$= 2 \arctan \sqrt{x} +c$$
A: Method 1. Algebraic substitution: 
let $\sqrt x=u\implies \frac{dx}{2\sqrt x}=du$ or $dx=2u\ du$, hence $$\int \frac{\sqrt x}{x^2+x}\ dx=\int \frac{u}{u^4+u^2}(2u\ du)=2\int \frac{1}{u^2+1}\ du=2\tan^{-1}(u)+C=2\tan^{-1}(\sqrt x)+C$$
Method 2. Trigonometric substitution: 
let $\sqrt x=\tan\alpha\implies \frac{dx}{2\sqrt x}=\sec^2\alpha\ d\alpha$ or $dx=2\tan\alpha\sec^2\alpha \ d\alpha$, hence $$\int \frac{\sqrt x}{x^2+x}\ dx=\int \frac{\tan\alpha}{\tan^4\alpha+tan^2\alpha}(2\tan\alpha\sec^2\alpha \ d\alpha)=2\int \frac{\tan^2\alpha\sec^2\alpha \ d\alpha}{\tan^2\alpha(\tan^2\alpha+1)}$$
$$=2\int \frac{\tan^2\alpha\sec^2\alpha \ d\alpha}{\tan^2\alpha\sec^2\alpha}=\int d\alpha=\alpha+C=2\tan^{-1}(\sqrt x)+C$$
