JEE Main 2021 — Mathematics Calculus

$$\int \frac{({x}^{2}-1)dx}{({x}^{4}+3{x}^{2}+1){\mathrm{tan}}^{-1}(x+\frac{1}{x})}+\int \frac{dx}{{x}^{4}+3{x}^{2}+1}$$∫(x4+3x2+1)tan−1(x+x1​)(x2−1)dx​+∫x4+3x2+1dx​

$$\int \frac{(1-\frac{1}{{x}^{2}})dx}{({(x+\frac{1}{x})}^{2}+1){\mathrm{tan}}^{-1}(x+\frac{1}{x})}+\frac{1}{2}\int \frac{({x}^{2}+1)-({x}^{2}-1)dx}{{x}^{4}+3{x}^{2}+1}$$∫((x+x1​)2+1)tan−1(x+x1​)(1−x21​)dx​+21​∫x4+3x2+1(x2+1)−(x2−1)dx​

Put ${\mathrm{tan}}^{-1}(x+\frac{1}{x})=t$tan−1(x+x1​)=t

$$\int \frac{dt}{t}+\frac{1}{2}\int \frac{(1+\frac{1}{{x}^{2}})dx}{{(x-\frac{1}{x})}^{2}+5}-\frac{1}{2}\int \frac{(1-\frac{1}{{x}^{2}})dx}{{(x+\frac{1}{x})}^{2}+1}$$∫tdt​+21​∫(x−x1​)2+5(1+x21​)dx​−21​∫(x+x1​)2+1(1−x21​)dx​

Put $x-\frac{1}{x}=y,x+\frac{1}{x}=z$x−x1​=y,x+x1​=z

$${\mathrm{log}}_{e}t+\frac{1}{2}\int \frac{dy}{{y}^{2}+5}-\frac{1}{2}\int \frac{dz}{{z}^{2}+1}$$loge​t+21​∫y2+5dy​−21​∫z2+1dz​

$$={\mathrm{log}}_{\text{e}}{\mathrm{tan}}^{-1}(x+\frac{1}{x})+\frac{1}{2\sqrt{5}}{\mathrm{tan}}^{-1}(\frac{{x}^{2}-1}{\sqrt{5}x})$$=loge​tan−1(x+x1​)+25​1​tan−1(5​xx2−1​)

$$-\frac{1}{2}{\mathrm{tan}}^{-1}(\frac{{x}^{2}+1}{x})+C$$−21​tan−1(xx2+1​)+C

$$\alpha =1,\beta =\frac{1}{2\sqrt{5}},\gamma =\frac{1}{\sqrt{5}},\delta =\frac{-1}{2}$$α=1,β=25​1​,γ=5​1​,δ=2−1​

or $\alpha =1,\beta =\frac{-1}{2\sqrt{5}},\gamma =\frac{-1}{\sqrt{5}},\delta =\frac{-1}{2}$α=1,β=25​−1​,γ=5​−1​,δ=2−1​

$$10(\alpha +\beta \gamma +\delta )=10(1+\frac{1}{10}-\frac{1}{2})=6$$10(α+βγ+δ)=10(1+101​−21​)=6