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I've solved this integral by partial integration method ('u' is first fraction and dv is the second fraction), but I've been told that there is a much simpler method using substitution which I can't see.I've tried to substitute arcsin(x) which eliminates the root and leaves me with sin functions in the integral (it didn't help). $$\int \frac {arcsin(x)}{\sqrt {1-x^2}} * \frac {1+x^2}{x^2} dx$$

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You're going to have to use integration by parts at some point. But as Harnak said, we can use $u$ substituion to have a much easier time. Let $u = \arcsin(x)$. Then \begin{align*} \int \frac{\arcsin(x)}{\sqrt{1-x^2}} (\frac{1+x^2}{x^2}) dx &= \int u(1+\csc^2(u))du \\ &=\frac{u^2}{2} +\int u\csc^2(u) du \\ &= \frac{u^2}{2} - u\cot(u) + \int \cot(u)du \\ &= \frac{u^2}{2} - u\cot(u)+\ln\mid\sin(u)\mid+C \end{align*} From this point, make the necessary substitutions back in and simplify.

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Hint: try the following substitution $$t= \arcsin(x)$$ and remember the derivative of arcsine.

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  • $\begingroup$ Are you sure it works ? $\endgroup$ – hamam_Abdallah Jan 30 at 17:30
  • $\begingroup$ It works according to wolframalpha.I've already tried this method, this gives me integral t* (1+sin^2 (t))/ sin^2 (t).I'm not sure how to proceed from here.Maybe trigonometric substitutions (t=tan(x/2))? $\endgroup$ – JoeDough Jan 30 at 17:56

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