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Question
Find `int_0^1 x(tan^-1x) "d"x`
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Solution
I = `int_0^1x(tan^-1x)^2 "d"x`
Integrating by parts, we have
I = `x^2/2[(tan^-1x)^2]_0^1 - 1/2 int_0^1 x^2 * 2 (tan^-1x)/(1 + x^2) "d"x`
= `pi^2/32 - int_0^1 x^2/(1 + x) * tan^-1 x"d"x`
= `pi^2/32 - 1_1`, where I1 = `int_0^1 x^2/(1 + x^2) tan^-1 x"d"x`
Now I1 = `int_0^1 (x^2 + 1 - 1)/(1 + x^2) tan^-1x "d"x`
= `int_0^1 tan^-1 x"d"x - int_0^1 1/(1 + x^2) tan^-1 x"d"x`
= `"I"_2 - 1/2 ((tan^-1x)^2)_0^1`
= `"I"_2 - pi^2/32`
Here I2 = `int_0^1 tan^-1 x"d"x = (x tan^-1x)_0^1 - int_0^1 x/(1 + x^2) "d"x`
= `pi/4 - 1/2(log|1 + x^2|)_0^1`
= `pi/4 - 1/2 log2`
Thus I2 = `pi/4 - 1/2 log 2 - pi^2/32`
Therefore, I = `pi^2/32 - pi/4 + 1/2 log2 + pi^2/32`
= `pi^2/16 - pi/4 + 1/2 log2`
= `(pi^2 - 4pi)/16 + log sqrt(2)`.
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