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Indefinite Integrals question

2021 · 16 Mar · Shift 2 · Q42
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  5. /2021 · 16 Mar · Shift 2 · Q42

Indefinite Integrals question

2021 · 16 Mar · Shift 2 · Q42

JEE MainMathematicsIndefinite IntegralsNumerical+4 / −1
For real numbers α\alphaα, β\betaβ, γ\gammaγ and δ\deltaδ, if ∫(x2−1)+tan⁡−1(x2+1x)(x4+3x2+1)tan⁡−1(x2+1x)dx=αlog⁡e(tan⁡−1(x2+1x))+βtan⁡−1(γ(x2+1)x)+δtan⁡−1(x2+1x)+C\int {{{({x^2} - 1) + {{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)} \over {({x^4} + 3{x^2} + 1){{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)}}dx} = \alpha {\log _e}\left( {{{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)} \right) + \beta {\tan ^{ - 1}}\left( {{{\gamma ({x^2} + 1)} \over x}} \right) + \delta {\tan ^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right) + C∫(x4+3x2+1)tan−1(xx2+1​)(x2−1)+tan−1(xx2+1​)​dx=αloge​(tan−1(xx2+1​))+βtan−1(xγ(x2+1)​)+δtan−1(xx2+1​)+C where C is an arbitrary constant, then the value of 10(α\alphaα+βγ\beta\gammaβγ+δ\deltaδ) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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