JEE Main 2023 — Mathematics Calculus

Given,

$$I=16{\int }_{1}^{2}\frac{dx}{{x}^{3}{({x}^{2}+2)}^{2}}$$I=16∫12​x3(x2+2)2dx​

$$\Rightarrow I=16{\int }_{1}^{2}\frac{dx}{{x}^{3}{x}^{4}{(1+\frac{2}{{x}^{2}})}^{2}}$$⇒I=16∫12​x3x4(1+x22​)2dx​

Now let $1+\frac{2}{{x}^{2}}=t\Rightarrow \frac{-4}{{x}^{3}}dx=dt$1+x22​=t⇒x3−4​dx=dt

Then, $\displaystyle I=-4{\int }_{3}^{\frac{3}{2}}\frac{dt}{{(\frac{2}{t-1})}^{2}{t}^{2}}$I=−4∫323​​(t−12​)2t2dt​

$$\Rightarrow I=-4{\int }_{3}^{\frac{3}{2}}{(\frac{t-1}{2})}^{2}\frac{dt}{{t}^{2}}$$⇒I=−4∫323​​(2t−1​)2t2dt​

$$\Rightarrow I=-\frac{4}{4}{\int }_{3}^{\frac{3}{2}}(1-\frac{2}{t}+\frac{1}{{t}^{2}})\mathrm{dt}$$⇒I=−44​∫323​​(1−t2​+t21​)dt

$$\Rightarrow I=-1{[t-2\mathrm{ln}|t|-\frac{1}{t}]}_{3}^{\frac{2}{2}}$$⇒I=−1[t−2ln∣t∣−t1​]322​​

$$\Rightarrow I=-1[(\frac{3}{2}-2\mathrm{ln}\frac{3}{2}-\frac{2}{3})-(3-2\mathrm{ln}3-\frac{1}{3})]$$⇒I=−1[(23​−2ln23​−32​)−(3−2ln3−31​)]

$$\Rightarrow I=-1[2\mathrm{ln}2-\frac{11}{6}]$$⇒I=−1[2ln2−611​]

$$\Rightarrow I=\frac{11}{6}-\mathrm{ln}4$$⇒I=611​−ln4