Find this integration

  • #1
utkarshakash
Gold Member
855
13

Homework Statement


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

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

[itex]\displaystyle \int_0^{\pi/4} tante^tdt [/itex]

Applying integration by parts by taking tant as 1st function

[itex]tant e^t - \displaystyle \int sec^2te^t dt [/itex]

This has made the problem more complicated instead of simplifying.
 

Answers and Replies

  • #2
3,812
92

Homework Statement


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

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

[itex]\displaystyle \int_0^{\pi/4} tante^tdt [/itex]

Applying integration by parts by taking tant as 1st function

[itex]tant e^t - \displaystyle \int sec^2te^t dt [/itex]

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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