Hello brilliant users, I'm Leonardo. I've some trouble to evaluate this integral. It took me several days to realize that this problem need some extraordinary way to solve it. Can somebody show me how to get the answer?
Thanks

The key is to use a substitution, e.g. \(u=\sqrt{x}\) which gives \(du=\dfrac1{2\sqrt{x}}\,dx\Leftrightarrow 2u\,du=dx\). Observe our integral becomes:\[\int\frac{\sqrt{x}}{1+x^2}dx=2\int\frac{u^2}{1+u^4}\,du\]next, factor using Germain's identity and decompose into partial fractions.

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## Comments

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TopNewestSubstituting \(x=u^{2}\) and we have, \(dx = 2udu\) Our integral becomes :

\[\Rightarrow \int \frac{\sqrt{x}}{1+x^{2}} dx = \int \frac{2u^{2}}{1+u^{4}} du\]

Adding and subtracting \(1\) in numerator,

\[\Rightarrow \int \frac{2u^{2}}{1+u^{4}}du\]

\[\Rightarrow \int \frac{(u^{2}+1)+(u^{2}-1)}{1+u^{4}}du\]

\[\Rightarrow \int \frac{u^{2}+1}{u^{4}+1}du + \int \frac{u^{2}-1}{u^{4}+1}du\]

Dividing by \(u^{2}\) in numerator and denominator,

\[\Rightarrow \int \frac{1+\frac{1}{u^{2}}}{u^{2}+ \frac{1}{u^{2}}}du + \int \frac{1-\frac{1}{u^{2}}}{u^{2}+ \frac{1}{u^{2}}}du\]

Making perfect square in denominator,

\[\Rightarrow \int \frac{1+\frac{1}{u^{2}}}{(u-\frac{1}{u})^{2}+2}du + \int \frac{1-\frac{1}{u^{2}}}{(u+\frac{1}{u})^{2}-2}du\]

Substituting \(u-\frac{1}{u} = t\) and \(u+\frac{1}{u} = v\) we get, \((1+\frac{1}{u^{2}})du = dt\) and similarly, \((1-\frac{1}{u^{2}})du = dv\)

Therefore, our integral becomes,

\[\Rightarrow \int \frac{dt}{t^{2}+(\sqrt{2})^{2}} + \int \frac{dv}{v^{2}-(\sqrt{2})^{2}}\]

\[\Rightarrow \frac{1}{\sqrt{2}} tan^{-1}(\frac{t}{\sqrt{2}}) + \frac{1}{2\sqrt{2}}\log _e \left( \frac{v-\sqrt{2}}{v+\sqrt{2}} \right) + C\]

Putting values of t and v, we get

\[\Rightarrow \frac{1}{\sqrt{2}} tan^{-1}(\frac{u-\frac{1}{u}}{\sqrt{2}}) + \frac{1}{2\sqrt{2}}\log _e \left( \frac{u+\frac{1}{u}-\sqrt{2}}{u+\frac{1}{u}+\sqrt{2}} \right) + C\]

Now, putting value of u, we get

\[\Rightarrow \frac{1}{\sqrt{2}} tan^{-1}(\frac{\sqrt{x}-\frac{1}{\sqrt{x}}}{\sqrt{2}}) + \frac{1}{2\sqrt{2}}\log _e \left( \frac{\sqrt{x}+\frac{1}{\sqrt{x}}-\sqrt{2}}{\sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{2}} \right) + C\]

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The key is to use a substitution, e.g. \(u=\sqrt{x}\) which gives \(du=\dfrac1{2\sqrt{x}}\,dx\Leftrightarrow 2u\,du=dx\). Observe our integral becomes:\[\int\frac{\sqrt{x}}{1+x^2}dx=2\int\frac{u^2}{1+u^4}\,du\]next, factor using Germain's identity and decompose into partial fractions.

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