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

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SUMMARY

The integral \(\int_0^1 \frac{xe^{\tan^{-1}x}}{\sqrt{1+x^2}} dx\) can be simplified using the substitution \(t = \tan^{-1}(x)\), which transforms the integral into \(\int_0^{\pi/4} \tan(t)e^t dt\). However, applying integration by parts with \(\tan(t)\) as the first function complicates the solution, leading to the expression \(\tan(t)e^t - \int \sec^2(t)e^t dt\). The derivative \(\frac{d}{dx}(\arctan(x))\) is confirmed to be \(\frac{1}{\sqrt{1+x^2}}\), necessitating careful handling of the substitution process.

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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}##.
 

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