I'm having a hard time doing this problem.
(Tan^-1(x)) / 1 + x^2
I'm know i'm gonna have to use the substitution rule
I designated
u = arctan
du = 1 / (1 + x^2) dx
From here i'm fairly stuck. Any help is appreciated
Hello,
You're right for the beginning !
$\displaystyle u=\arctan(x)$
$\displaystyle du=\frac{1}{1+x^2} dx \rightarrow dx=(1+x^2)du$
Then $\displaystyle \int \frac{\arctan(x)}{1+x^2} dx=\int \frac{u}{1+x^2} dx=\int \frac{u}{\color{red}1+x^2} ({\color{red}1+x^2}) du$$\displaystyle =\int udu$
??? You're stuck? You just substitute it in ...
$\displaystyle {\color{red} u = \arctan x}$
$\displaystyle {\color{blue} du = \frac{1}{1 +x^{2}} dx}$
Looking at your integral:
$\displaystyle \int \frac{{\color{red}\arctan x}}{{\color{blue}1 + x^{2}}} {\color{blue} dx} = \int \left({\color{red}\arctan x} \cdot {\color{blue}\frac{1}{1+x^{2}}dx}\right) = \int {\color{red} u} {\color{blue}du}$