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Question
Evaluate: `int_0^1 x* tan^-1x "d"x`
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Solution
Let I = `int_0^1 x tan^-1x "d"x`
= `[tan^-1 x int x "d"x]_0^1 - int_0^1["d"/("d"x)(tan^-1x) int x "d"x]"d"x`
= `[tan^-1x* x^2/2]_0^1 -int_0^1 1/(1 + x^2)*x^2/2 "d"x`
= `[x^2/2 tan^-1x]_0^1 - 1/2 int_0^1 x^2/(1 + x^2) "d"x`
= `[1/2 tan^-1 - 0] - 1/2 int (x^2 + 1 - 1)/(1 + x^2) "d"x`
= `1/2* pi/4 - 1/2 int_0^1 (1 - 1/(1 + x^2)) "d"x`
= `pi/8 - 1/2[x - tan^-1x]_0^1`
= `pi/8 - 1/2[(1 - tan^-1 1) - (0 - tan^-1 0)]`
= `pi/8 - 1/2(1 - pi/4 - 0)`
= `pi/8 - 1/2 + pi/8`
∴ I = `pi/4 - 1/2`
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