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← JEE Main 2022 · 28 Jul · Shift 1
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JEE Main 2022 · 28 Jul · Shift 1, Mathematics question 24

If ∫0315x31+x2+(1+x2)3dx=α2+β3\displaystyle\int_0^{\sqrt3}\frac{15x^3}{\sqrt{1+x^2+\sqrt{\left(1+x^2\right)^3}}}dx=\alpha\sqrt2+\beta\sqrt3, where α\alpha, β\beta are integers, then α+β\alpha+\beta is equal to ______.

Answer and worked solution

Official answer: 10

Simplify. 1+x2+(1+x2)3/2=(1+x2)(1+1+x2)1+x^2+(1+x^2)^{3/2}=(1+x^2)\left(1+\sqrt{1+x^2}\right). With u=1+x2u=\sqrt{1+x^2} the integral becomes 15∫12u2−11+udu15\int_1^2\frac{u^2-1}{\sqrt{1+u}}du.

Evaluate. It comes to 162−6316\sqrt2-6\sqrt3, so α+β=10\alpha+\beta=10.

Answer: 10

Topics:Properties of definite integralsSubstitution
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