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प्रश्न
Evaluate the following : `int_0^1 (log(x + 1))/(x^2 + 1)*dx`
उत्तर
Let I = `int_0^1 (log(x + 1))/(x^2 + 1)*dx`
Put x = tan θ.
∴ dx = sec2θ·dθ
and
x2 + 1 = tan2θ + 1 = sec2θ
When x = 0, tan θ = 0 ∴ θ = 0
When x = 1, tan θ = 1 ∴ θ = `pi/(4)`
∴ I = `int_0^(pi/4) (log(tan theta + 1))/sec^2 theta* sec2 theta *d theta`
= `int_0^(pi/4) log(1 + tan theta)*d theta` ...(1)
We use the property, `int_0^a f(x)*dx = int_0^a f(a - x)*dx`.
Here, a = ``pi/(4).
Hence changing θ by `pi/(4) - theta`, we have,
I = `int_0^(pi/4) log[1 + tan(pi/4 - 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) [log 2 - log (1 + tan theta)]*d theta`
= `log 2 int_0^(pi/4) 1*d theta - int_0^(pi/4) log(1 + tan theta)*d theta`
= `(log 2) [theta]_0^(pi/4) - "I"`
= `pi/(4) log 2 - "I"`
∴ 2I = `pi/(4) log 2`
∴ I = `pi/(8) log 2`.
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