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

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The integral in question is ∫₀¹ (xe^(tan⁻¹x)/√(1+x²)) dx, which is approached through the substitution tan⁻¹(x) = t, leading to x = tan(t) and dt = dx/√(1+x²). This transforms the integral into ∫₀^(π/4) (tan(t)e^t) dt. However, applying integration by parts complicates the problem, resulting in an expression that is more complex rather than simplified. A suggestion is made to multiply and divide by √(1+x²) to facilitate the substitution.
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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}##.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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