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Question

Solve 0log(x+1x)dx1+x2

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Solution

0log(x+1x)dx1+x2

Put x=tanθ
dx=sec2θdθ
When x=0θ=0
When x=θ=π2

I=π/20log(tanθ+cotθ)dθ

=π/20logsin2θ+cos2θsinθcosθdθ

=π/20logsinθcosθdθ

We know that
π/20logsinθ=π2log2
π/20logcosθ=π2log2

=[π/20logsinθdθ+π/20logcosθdθ]

=[π2log2π2log2]

I=πlog2.

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