JEE Main 2021 — Mathematics Calculus

Let $\sqrt{x}=t$x​=t

$$\Rightarrow x={t}^{2},dx=2tdt$$⇒x=t2,dx=2tdt

Let $\displaystyle I={\int }_{0}^{1}\frac{\sqrt{x}dx}{(1+x)(1+3x)(3+x)}$I=∫01​(1+x)(1+3x)(3+x)x​dx​

$$I={\int }_{0}^{1}\frac{2{t}^{2}dt}{({t}^{2}+1)(3{t}^{2}+1)({t}^{2}+3)}$$I=∫01​(t2+1)(3t2+1)(t2+3)2t2dt​

$$={\int }_{0}^{1}\frac{((3{t}^{2}+1)-({t}^{2}+1))dt}{({t}^{2}+1)(3{t}^{2}+1)({t}^{2}+3)}$$=∫01​(t2+1)(3t2+1)(t2+3)((3t2+1)−(t2+1))dt​

$$={\int }_{0}^{1}(\frac{1}{({t}^{2}+3)({t}^{2}+1)}-\frac{1}{({t}^{2}+3)(3{t}^{2}+1)})dt$$=∫01​((t2+3)(t2+1)1​−(t2+3)(3t2+1)1​)dt

$$={\int }_{0}^{1}(\frac{dt}{2({t}^{2}+1)}-\frac{1}{8}\frac{3dt}{(3{t}^{2}+1)}-\frac{3}{8}\frac{dt}{({t}^{2}+3)})$$=∫01​(2(t2+1)dt​−81​(3t2+1)3dt​−83​(t2+3)dt​)

$$={(\frac{1}{2}{\mathrm{tan}}^{-1}⁡t-\frac{3\sqrt{3}}{8\times 3}{\mathrm{tan}}^{-1}⁡\sqrt{3}t-\frac{3}{8\sqrt{3}}{\mathrm{tan}}^{-1}⁡\frac{t}{\sqrt{3}})}_{0}^{1}$$=(21​tan−1⁡t−8×333​​tan−1⁡3​t−83​3​tan−1⁡3​t​)01​

$$=\frac{\pi }{8}-\frac{\sqrt{3}}{8}\times \frac{\pi }{3}-\frac{\sqrt{3}}{8}\times \frac{\pi }{6}$$=8π​−83​​×3π​−83​​×6π​

$$=\frac{\pi }{8}-\frac{\sqrt{3}\pi }{16}$$=8π​−163​π​

$$=\frac{\pi }{8}(1-\frac{\sqrt{3}}{2})$$=8π​(1−23​​)