Question: $\color{green}{question}$:

How do you evaluate this integral?

> 
> $$\int \frac{y^2}{y^2+d^2}dy=y-d\,{\tan}^{-1}\left ( \frac{y}{d} \right )+\mathrm{constant}$$
> 
> 
> 

$\color{green}{I~know}$ I should use the change of variables, But I do not know how to do.

Thank you for any hint.

Answer: 
**Hint**

* $$\frac{y^2}{y^2+d^2}=\frac{y^2+d^2-d^2}{y^2+d^2}=1-\frac{d^2}{y^2+d^2}$$
* $$\frac{d^2}{y^2+d^2}=\frac{1}{(y/d)^2+1}=\frac{1}{t^2+1}$$
* $$\int\frac{dt}{t^2+1}=\arctan t+C$$
Maple code with(Student[Calculus1]):IntTutor$(y^2/(y^2 +d^2),y$); produced the output which can be seen [here](http://rapidshare.com/files/1020936765/screen2.24.08.13.docx). It coincides with the excellent answer by Eric Auld.