JEE Main 2024 — Mathematics Calculus

Let, $\displaystyle I=\int \frac{{x}^{8}-{x}^{2}}{({x}^{12}+3{x}^{6}+1){\mathrm{tan}}^{-1}({x}^{3}+\frac{1}{{x}^{3}})}\mathrm{dx}$I=∫(x12+3x6+1)tan−1(x3+x31​)x8−x2​dx

Putting, ${\mathrm{tan}}^{-1}({x}^{3}+\frac{1}{{x}^{3}})=t$tan−1(x3+x31​)=t

$$\Rightarrow \frac{1}{1+{({x}^{3}+\frac{1}{{x}^{3}})}^{2}}\cdot (3{x}^{2}-\frac{3}{{x}^{4}})\mathrm{dx}=\mathrm{dt}$$⇒1+(x3+x31​)21​⋅(3x2−x43​)dx=dt

$$\Rightarrow \frac{1}{1+({x}^{6}+\frac{1}{{x}^{6}}+2)}\cdot (3{x}^{2}-\frac{3}{{x}^{4}})\mathrm{dx}=\mathrm{dt}$$⇒1+(x6+x61​+2)1​⋅(3x2−x43​)dx=dt

$$\Rightarrow \frac{{x}^{6}}{{x}^{12}+3{x}^{6}+1}\cdot \frac{3{x}^{6}-3}{{x}^{4}}\mathrm{dx}=\mathrm{dt}$$⇒x12+3x6+1x6​⋅x43x6−3​dx=dt

$$\Rightarrow \frac{({x}^{8}-{x}^{2})}{{x}^{12}+3{x}^{6}+1}\mathrm{dx}=\frac{\mathrm{dt}}{3}$$⇒x12+3x6+1(x8−x2)​dx=3dt​

$$\Rightarrow I=\frac{1}{3}\int \frac{1}{t}\mathrm{dt}$$⇒I=31​∫t1​dt

$$\Rightarrow I=\frac{1}{3}\mathrm{log}|t|+C$$⇒I=31​log∣t∣+C

$$\Rightarrow I=\frac{1}{3}\mathrm{log}|{\mathrm{tan}}^{-1}({x}^{3}+\frac{1}{{x}^{3}})|+C$$⇒I=31​log∣tan−1(x3+x31​)∣+C

$$\Rightarrow I=\mathrm{log}{|{\mathrm{tan}}^{-1}({x}^{3}+\frac{1}{{x}^{3}})|}^{\frac{1}{3}}+C$$⇒I=log∣tan−1(x3+x31​)∣31​+C

Hence optin $(A)$(A) is correct