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Question Number 86957 by abdomathmax last updated on 01/Apr/20

$$find\int arctan\left(\frac{1-u}{1+u}\right)du$$

Commented by Ar Brandon last updated on 01/Apr/20

$$I=\int arctan\left(\frac{1-x}{1+x}\right)dx$$$$Letu=arctan\left(\frac{1-x}{1+x}\right)\iff u=arctan\left(t\right)witht=\frac{1-x}{1+x}$$$$\Rightarrow \frac{du}{dx}=\frac{du}{dt}\times \frac{dt}{du}$$$$\frac{du}{dt}=\frac{1}{1+t^2}\frac{dt}{du}=\frac{-(1+x)-(1-x)}{{(1+x)}^2}=\frac{-2}{{(1+x)}^2}$$$$\Rightarrow \frac{du}{dx}=\frac{1}{1+{\left(\frac{1-x}{1+x}\right)}^2}\times \frac{-2}{{(1+x)}^2}=\frac{{(1+x)}^2}{{(1+x)}^2+{(1-x)}^2}\times \frac{-2}{{(1+x)}^2}=\frac{-2}{(1+2x+x^2)+(1-2x+x^2)}$$$$\Rightarrow \frac{du}{dx}=-\frac{1}{1+x^2}\Rightarrow du=-\frac{1}{1+x^2}dx$$$$$$$$Letalsodv=dx\Rightarrow v=x$$$$$$$$ThusI=xarctan\left(\frac{1-x}{1+x}\right)+\int \frac{x}{1+x^2}dx$$$$=xarctan\left(\frac{1-x}{1+x}\right)+\frac{1}{2}ln(1+x^2)+C$$$$\int arctan\left(\frac{1-u}{1+u}\right)du=uarctan\left(\frac{1-u}{1+u}\right)+\frac{1}{2}ln(1+u^2)+C$$$$C\in \mathbb{R}$$$$$$$$$$

Commented by mathmax by abdo last updated on 01/Apr/20

$$I=\int arctan\left(\frac{1-u}{1+u}\right)dubyparts{f}^{{}^{\prime }}=1andv=arctan\left(\frac{1-u}{1+u}\right)\Rightarrow$$$$v^{{}^{\prime }}=\frac{{\left(\frac{1-u}{1+u}\right)}^{{}^{\prime }}}{1+{\left(\frac{1-u}{1+u}\right)}^2}=\frac{-(1+u)-(1-u)}{{(1+u)}^2}\times \frac{{(1+u)}^2}{{(1+u)}^2+{(1-u)}^2}$$$$=\frac{-2}{u^2+2u+1+u^2-2u+1}=\frac{-2}{2u^2+2}=-\frac{1}{u^2+1}\Rightarrow$$$$I=uarctan\left(\frac{1-u}{1+u}\right)+\int \frac{u}{u^2+1}du+c$$$$=uarctan\left(\frac{1-u}{1+u}\right)+\frac{1}{2}ln(1+u^2)+C$$$$$$

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