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Question Number 63261 by aliesam last updated on 01/Jul/19

$$\int xtan\left(x\right)dx$$

Commented by kaivan.ahmadi last updated on 01/Jul/19

$$u=tanx\Rightarrow du=\frac{dx}{1+x^2}$$$$dv=xdx\Rightarrow v=\frac{x^2}{2}$$$$uv-\int vdu=\frac{x^2}{2}tanx-\frac{1}{2}\int \frac{x^2}{1+x^2}dx=$$$$\frac{x^2}{2}tanx-\frac{1}{2}\int (1-\frac{1}{1+x^2})dx=$$$$\frac{x^2}{2}tanx-\frac{1}{2}(x-tanx)+C$$

Commented by aliesam last updated on 01/Jul/19

$$welldonesir$$

Commented by aliesam last updated on 01/Jul/19

$$noproblemandnicesolutionsirthanks$$

Commented by mathmax by abdo last updated on 01/Jul/19

$$letA=\int xtan\left(x\right)dxchangementtanx=tgive$$$$A=\int tarctan\left(t\right)byparts{u}^{{}^{\prime }}=tandv=arctan\left(t\right)$$$$A=\frac{t^2}{2}arctan\left(t\right)-\int \frac{t^2}{2}\frac{dt}{1+t^2}+c=\frac{t^2}{2}arctan\left(t\right)-\frac{1}{2}\int \frac{1+t^2-1}{1+t^2}dt+c$$$$=\frac{t^2}{2}arctan\left(t\right)-\frac{t}{2}-\frac{arctan\left(t\right)}{2}+c\Rightarrow$$$$A=\frac{x{\left(arctanx\right)}^2}{2}-\frac{tanx}{2}-\frac{x}{2}+c.$$

Commented by mathmax by abdo last updated on 01/Jul/19

$$sorryA=\frac{x}{2}{tan}^{2}x-\frac{tanx}{2}-\frac{x}{2}+c.$$

Commented by mathmax by abdo last updated on 02/Jul/19

$$youarewelcome.$$

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