What is `int_0^1 ` e^(arctgx)/(x^2+1)^(3/2)dx?

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You need to re-write the integrand such that:

`int_0^1 e^(arctanx)/(x^2+1)^(3/2)dx = int_0^1 (e^(arctanx))/((x^2+1)sqrt(x^2 + 1)) dx`

You should notice that `(e^(arctanx))' = (e^(arctanx))/(x^2+1)` , hence, you may replace `(e^(arctanx))'` for `(e^(arctanx))/(x^2+1)` , such that:

`int_0^1 (e^(arctanx))/((x^2+1)sqrt(x^2 + 1)) dx = int_0^1 ((e^(arctanx))')/(sqrt(x^2 + 1)) dx`

You should use integration by parts, such that:

`int f(x)*g'(x)dx = f(x)*g(x) - int f'(x)*g(x)dx`

Let `1/(sqrt(x^2 + 1)) = f(x)` and `((e^(arctanx))') = g'(x)` , such that:

`f'(x) = -x/((x^2+1)sqrt(x^2 + 1))`

`g(x) = e^(arctanx)`

Using the formula yields:

`int_0^1 ((e^(arctanx))')/(sqrt(x^2 + 1)) dx = (e^(arctanx))/(sqrt(x^2 + 1))|_0^1 + int_0^1 (xe^(arctanx))/((x^2+1)sqrt(x^2 + 1)) dx`

`int_0^1 ((e^(arctanx))')/(sqrt(x^2 + 1)) dx = (e^(arctanx))/(sqrt(x^2 + 1))|_0^1 + int_0^1 ((e^(arctanx))')*x/(sqrt(x^2 + 1)) dx`

You need to solve the integral `int_0^1 ((e^(arctanx))')*x/(sqrt(x^2 + 1))dx` using parts again, such that:

`f(x) = x/(sqrt(x^2 + 1)) => f'(x) = (x^2 + 1 - x^2)/((x^2 + 1)sqrt(x^2 + 1))`

`f'(x) = 1/((x^2 + 1)sqrt(x^2 + 1))`

`g'(x) = ((e^(arctanx))') => g(x) = e^(arctanx)`

`int_0^1 ((e^(arctanx))')*x/(sqrt(x^2 + 1))dx = (xe^(arctanx))/(sqrt(x^2 + 1))|_0^1 - int_0^1 (e^(arctanx))/((x^2 + 1)sqrt(x^2 + 1)) dx`

Let `I = int_0^1 (e^(arctanx))/((x^2 + 1)sqrt(x^2 + 1)) dx` , hence, evaluating `I` , yields:

`I = (e^(arctanx))/(sqrt(x^2 + 1))|_0^1 + (xe^(arctanx))/(sqrt(x^2 + 1))|_0^1 - I`

Moving the terms that contain I to the left side, yields:

`2I = (e^(arctanx))/(sqrt(x^2 + 1))|_0^1 + (xe^(arctanx))/(sqrt(x^2 + 1))|_0^1`

By fundamental theorem of calculus, yields:

`2I = e^(arctan 1)/sqrt 2 - e^0/sqrt 1 + e^(arctan 1)/sqrt 2 - 0`

`2I = e^(pi/4)/sqrt2 - 1 + e^(pi/4)/sqrt2`

`I = e^(pi/4)/sqrt2 - 1/2`

Hence, evaluating the given definite integral, using integration by parts, yields `I = e^(pi/4)/sqrt2 - 1/2` .

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