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Evaluate the integral of (x earctan x)/(1+x2)3/2

by RoRi

Evaluate the following integral

    \[ \int \frac{x e^{\arctan x}}{(1+x^2)^{\frac{3}{2}}} \, dx. \]

This will take two integration by parts steps. For the first one let

    \begin{align*} u &= e^{\arctan x} & du &= \frac{e^{\arctan x}}{1+x^2} \, dx \\ dv &= \frac{x}{(1+x^2)^{\frac{3}{2}}} \, dx & v &= \frac{-1}{\sqrt{1+x^2}}. \end{align*}

Then we have

    \begin{align*} \int \frac{x e^{\arctan x}}{(1+x^2)^{\frac{3}{2}}} \, dx &= -\frac{e^{\arctan x}}{\sqrt{1+x^2}} + \int \frac{e^{\arctan x}}{(1+x^2)^{\frac{3}{2}}} \, dx. \end{align*}

Now, for the integral on the left we integrate by parts again. This time let

    \begin{align*} u &= e^{\arctan x} & du &= \frac{e^{\arctan x}}{1+x^2} \, dx \\ dv &= \frac{1}{(1+x^2)^{\frac{3}{2}}} \, dx & v &= \frac{x}{\sqrt{1+x^2}}. \end{align*}

So for this integral we have,

    \[ \int \frac{e^{\arctan x}}{(1+x^2)^{\frac{3}{2}}} \, dx = \frac{x e^{\arctan x}}{\sqrt{1+x^2}} - \int \frac{x e^{\arctan x}}{(1+x^2)^{\frac{3}{2}}} \, dx.\]

Plugging this back into our formula for the integral we are trying to evaluate we have,

    \begin{align*} && \int \frac{x e^{\arctan x}}{(1+x^2)^{\frac{3}{2}}} \, dx &= -\frac{e^{\arctan x}}{\sqrt{1+x^2}} + \int \frac{e^{\arctan x}}{(1+x^2)^{\frac{3}{2}}} \, dx \\[10pt] \implies && \int \frac{x e^{\arctan x}}{(1+x^2)^{\frac{3}{2}}} \, dx &= -\frac{e^{\arctan x}}{\sqrt{1+x^2}} + \frac{x e^{\arctan x}}{\sqrt{1+x^2}} - \int \frac{x e^{\arctan x}}{(1+x^2)^{\frac{3}{2}}} \, dx \\[10pt] \implies && 2 \int \frac{x^{\arctan x}}{(1+x^2)^{\frac{3}{2}}} \, dx &= -\frac{e^{\arctan x}}{\sqrt{1+x^2}} + \frac{x e^{\arctan x}}{(1+x^2)^{\frac{3}{2}}} + C \\[10pt] \implies && \int \frac{x^{\arctan x}}{(1+x^2)^{\frac{3}{2}}} \, dx &= \frac{(x-1) e^{\arctan x}}{2 \sqrt{1+x^2}} + C. \end{align*}

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