JEE Main 2023 — Mathematics Calculus

Given,

$$I(x)=\int \sqrt{\frac{x+7}{x}}dx$$I(x)=∫xx+7​​dx

Now let, $\frac{x+7}{x}={t}^{2}\Rightarrow -\frac{7}{{x}^{2}}dx=2tdt$xx+7​=t2⇒−x27​dx=2tdt

$$\Rightarrow dx=\frac{-14t}{{({t}^{2}-1)}^{2}}dt$$⇒dx=(t2−1)2−14t​dt

So, $\displaystyle I(x)=-14\int \frac{{t}^{2}}{{({t}^{2}-1)}^{2}}dt$I(x)=−14∫(t2−1)2t2​dt

$$\Rightarrow I(x)=-14\int \frac{dt}{({t}^{2}+\frac{1}{{t}^{2}}-2)}$$⇒I(x)=−14∫(t2+t21​−2)dt​

$$\Rightarrow I(x)=\frac{-14}{2}\int [\frac{(1-\frac{1}{{t}^{2}})}{{(t+\frac{1}{t})}^{2}-4}+\frac{(1+\frac{1}{{t}^{2}})}{{(t-\frac{1}{t})}^{2}}]dt$$⇒I(x)=2−14​∫[(t+t1​)2−4(1−t21​)​+(t−t1​)2(1+t21​)​]dt

$$\Rightarrow I(x)=-7(\frac{1}{4}\mathrm{ln}|\frac{t+\frac{1}{t}-2}{t+\frac{1}{t}+2}|-\frac{1}{t-\frac{1}{t}})+c$$⇒I(x)=−7(41​ln∣t+t1​+2t+t1​−2​∣−t−t1​1​)+c

Now when $x=9,t=\frac{4}{3}$x=9,t=34​

$$\Rightarrow I(9)=12+7\times \mathrm{ln}7=\frac{-7}{4}\mathrm{ln}{(\frac{1}{7})}^{2}+7\times \frac{12}{7}+c$$⇒I(9)=12+7×ln7=4−7​ln(71​)2+7×712​+c

$$\Rightarrow c=\frac{7}{2}\mathrm{ln}7$$⇒c=27​ln7

Now when $x=1,t=2\sqrt{2}$x=1,t=22​

$$\Rightarrow I(1)=+\frac{7}{4}\mathrm{ln}{(\frac{2\sqrt{2}+1}{2\sqrt{2}-1})}^{2}+7\times \frac{2\sqrt{2}}{7}+\frac{7}{2}\mathrm{ln}7$$⇒I(1)=+47​ln(22​−122​+1​)2+7×722​​+27​ln7

$$\Rightarrow I(1)=\frac{7}{2}\mathrm{ln}(\frac{{(2\sqrt{2}+1)}^{2}}{7})+2\sqrt{2}+\frac{7}{2}\mathrm{ln}7$$⇒I(1)=27​ln(7(22​+1)2​)+22​+27​ln7

$$\Rightarrow I(1)=7\mathrm{ln}(2\sqrt{2}+1)-\frac{7}{2}\mathrm{ln}7+2\sqrt{2}+\frac{7}{2}\mathrm{ln}7$$⇒I(1)=7ln(22​+1)−27​ln7+22​+27​ln7

$$\Rightarrow \alpha =2\sqrt{2}\Rightarrow {\alpha }^{4}=64$$⇒α=22​⇒α4=64