Find this Integration: Is There a Simplified Form for This Integral?

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Homework Statement


\displaystyle \int_0^1 \dfrac{xe^{tan^{-1}x}}{\sqrt{1+x^2}} dx

Homework Equations



The Attempt at a Solution


Let tan^-1 (x) = t
x = tant
dt=dx/sqrt{1+x^2}
The integral then reduces to

\displaystyle \int_0^{\pi/4} tante^tdt

Applying integration by parts by taking tant as 1st function

tant e^t - \displaystyle \int sec^2te^t dt

This has made the problem more complicated instead of simplifying.
 
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utkarshakash said:

Homework Statement


\displaystyle \int_0^1 \dfrac{xe^{tan^{-1}x}}{\sqrt{1+x^2}} dx

Homework Equations



The Attempt at a Solution


Let tan^-1 (x) = t
x = tant
dt=dx/sqrt{1+x^2}
The integral then reduces to

\displaystyle \int_0^{\pi/4} tante^tdt

Applying integration by parts by taking tant as 1st function

tant e^t - \displaystyle \int sec^2te^t dt

This has made the problem more complicated instead of simplifying.

$$\frac{d}{dx}\left(\arctan(x)\right) ≠ \frac{1}{\sqrt{1+x^2}}$$
To use the substitution, multiply and divide by ##\sqrt{1+x^2}##.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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