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**Integral: Hint plz**

[tex]\int\frac{x\tan^{-1}x}{(1+x^2)^3}dx[/tex]

[tex]\int\frac{\tan^{ - 1}x}{(1 + x^2)}\frac{x}{(1 + x^2)^2}dx\left\{\begin{array}{cc}x = \tan{r} \leftrightarrow r = \tan^{ - 1}x \\

dr = \frac {dx}{1 + x^2}\end{array}\right\}[/tex]

Let's not even get into what I did next and how much work I put into this Integral. This is definitely not the right way to go, hmm ...

Ok I think I see it now ... omg.

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