The value of the integral \\int_{1} ^{\\sqrt{2}+1}({x^{2}-1}/{x^{2}+1}) {1}/{\\sqrt{1+x^{4}} d x is

KVPY 2020-21 SB SX Exam Previous Papers

Question : 89 of 120

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The value of the integral

\∫↙{1} ↖{\√{2}+1}({x^{2}-1}/{x^{2}+1}) {1}/{\√{1+x^{4}}} d x

Solution:

∫↙{1} ↖{\√{2}+1} {(x^{2}-1)}/{x(x+{1}/{x}) x \√{x^{2}+{1}/{x^{2}}}} d x

=∫↙{1} ↖{\√{2}+1} {1-{1}/{x^{2}}}/{(x+{1}/{x}) \√{(x+{1}/{x})^{2}-2}} {dx}

Let

x+{1}/{x}=\√{2} \sec \θ

(1-{1}/{x^{2}}) d x=\√{2} \sec \θ\tan \θd \θ

\∫↙{\π\/ 4} ↖{\π\/ 3} {\√{2} \sec \θ\tan \θd \θ}/{\√{2} \sec \θ\β{2} \tan \θ}

={\π}/{12 \√{2}}

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