JEE Main 2023 — Mathematics Calculus

Given,

$$I(x)=\int \frac{x+1}{x{(1+x{e}^{x})}^{2}}dx$$I(x)=∫x(1+xex)2x+1​dx

Now let $1+x{e}^{x}=t$1+xex=t

$$\Rightarrow {e}^{x}(x+1)dx=dt$$⇒ex(x+1)dx=dt

So, $\displaystyle I(x)=\int \frac{1}{(t-1){t}^{2}}dt$I(x)=∫(t−1)t21​dt

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

$$\Rightarrow I(x)=\int \frac{-(t+1)}{{t}^{2}}+\frac{1}{t-1}dt$$⇒I(x)=∫t2−(t+1)​+t−11​dt

$$\Rightarrow I(x)=\int -\frac{1}{t}-\frac{1}{{t}^{2}}+\frac{1}{t-1}dt$$⇒I(x)=∫−t1​−t21​+t−11​dt

$$\Rightarrow I(x)=-\mathrm{ln}t+\frac{1}{t}+\mathrm{ln}(t-1)+C$$⇒I(x)=−lnt+t1​+ln(t−1)+C

$$\Rightarrow I(x)=\mathrm{ln}(\frac{x{e}^{x}}{x{e}^{x}+1})+\frac{1}{x{e}^{x}+1}+C$$⇒I(x)=ln(xex+1xex​)+xex+11​+C

Also given,$\underset{x\rightarrow \infty }{\mathrm{lim}}I(x)=0$x→∞lim​I(x)=0

$$\Rightarrow \underset{x\rightarrow \infty }{\mathrm{lim}}I(x)=\underset{x\rightarrow \infty }{\mathrm{lim}}[\mathrm{ln}(1-\frac{1}{x{e}^{x}+1})+\frac{1}{x{e}^{x}+1}+C]$$⇒x→∞lim​I(x)=x→∞lim​[ln(1−xex+11​)+xex+11​+C]

$$\Rightarrow \underset{x\rightarrow \infty }{\mathrm{lim}}I(x)=[\mathrm{ln}(1-0)+0+C]$$⇒x→∞lim​I(x)=[ln(1−0)+0+C]

$$\Rightarrow C=0$$⇒C=0

Now finding,

$$I(1)=\mathrm{ln}(\frac{e}{e+1})+\frac{1}{e+1}=1+\frac{1}{e+1}-\mathrm{ln}(e+1)$$I(1)=ln(e+1e​)+e+11​=1+e+11​−ln(e+1)

$$\Rightarrow I(1)=\frac{e+2}{e+1}-\mathrm{ln}(e+1)$$⇒I(1)=e+1e+2​−ln(e+1)