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Question
Evaluate: `int_0^1 (log(x + 1))/(x^2 + 1) "d"x`
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Solution
Let I = `int_0^1 (log(x + 1))/(x^2 + 1) "d"x`
Put x = tan θ
∴ dx = sec2θ dθ
When x = 0, θ = 0 and when x = 1, θ = `pi/4`
∴ I = `int_0^(pi/4)(log(tantheta + 1))/(tan^2theta + 1) xx sec^2theta "d"theta`
= `int_0^(pi/4) (log(1 + tantheta))/(sec^2theta) xx sec^2theta "d"theta`
∴ I = `int_0^(pi/4) log(1 + tan theta) "d"theta` ......(i)
= `int_0^(pi/4) log[1 + tan(pi/4 - theta)] "d"theta` ......`[∵ int_0^"a" "f"(x) "d"x = int_0^"a" "f"("a" - x) "d"x]`
= `int_0^(pi/4) log[1 + (tan pi/4 - tantheta)/(1 + tan pi/4 tan theta)] "d"theta`
= `int_0^(pi/4) log[1 + (1 - tan theta)/(1 + tan theta)] "d"theta`
= `int_0^(pi/4) log[(1 + tan theta + 1 - tan theta)/(1 + tan theta)] "d"theta`
= `int_0^(pi/4) log[2/(1 + tan theta)] "d"theta`
= `int_0^(pi/4) [log2 - log(1 + tan theta)] "d"theta`
= `log 2 int_0^(pi/4) "d"theta - int_0^(pi/4) log(1 + tan theta) "d"theta`
∴ I = `log 2[theta]_0^(pi/4) - "I"` .....[From (i)]
∴ 2I = `log 2(pi/4 - 0)`
∴ I = `pi/8 log 2`
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