Other method Indefinite Integral :$\int \frac{1}{1+x^2} dx$

Question

Question

How indefinite integral : $$\int \frac{1}{1+x^2} dx$$

I know this $$\int \frac{dx}{1+x^2}=\arctan x + C$$

One of my friends in high school studied integration and came across this example. He does not know arctan x and does not know anything about it. I want to help him answer this question. My question is, are there other ways to answer this integral?

Answer 1

By partial fractions, $$\begin{array} .\int\frac{dx}{x^2+1}&=\frac{1}{2i}\left(\frac1{x-i}-\frac1{x+i}\right)dx\\ &=\frac1{2i}(\ln(x-i)-\ln(x+i))+C\\ &=\frac1{2i}\ln(\tfrac{x-i}{x+i})+C \end{array}$$

Answer 2

Just horsing around: $$\int \frac{dx}{1+x^2} = \left\{x = \sinh t\right\} = \int \frac{\cosh t}{\cosh^2t}dt = \int\frac{dt}{\cosh t}=$$ $$\int\sum_{n=0}^{+\infty}\frac{E_nt^n}{n!}dt = \sum_{n=0}^{+\infty}\frac{E_n}{n!}\int t^n dt=C+\sum_{n=0}^{+\infty}\frac{E_nt^{n+1}}{(n+1)!} = $$ $$=C+\sum_{n=1}^{+\infty} \frac{E_{n-1}t^n}{n!}=C + \sum_{n=1}^{+\infty}\frac{E_{n-1}\sinh^nx}{n!}$$

Answer 3

Substitute $x=\tan t$. Then, $dx = \sec^2 t\ dt$ $$\int\frac1{1+x^2}dx = \int \frac {\sec^2t}{1+\tan^2t}dt = \int dt =t +C$$