Question

If $p\in R$ and $C$ is arbitrary constant of integration, then $\int \frac{(e^x+1)dx}{e^{2x}+x^2+2x{e}^x-p^3-1}$ is equal to

• A. $\frac{1}{\sqrt{-p^3-1}}{\tan }^{-1}\left(\frac{e^x+x}{\sqrt{-p^3-1}}\right)+C$ for $p<-1$
• B. $\frac{1}{2\sqrt{-p^3-1}}\ln \left|\frac{e^x+x-\sqrt{-p^3-1}}{e^x+x+\sqrt{-p^3-1}}\right|+C$ for $p<-1$
• C. $\frac{-1}{e^x+x}+C$ for $p=-1$
• D. $\frac{1}{2\sqrt{p^3+1}}\ln \left|\frac{e^x+x-\sqrt{p^3+1}}{e^x+x+\sqrt{p^3+1}}\right|+C$ for $p>-1$

Answer 1

Answer: (A), (C), (D)

$\frac{1}{2\sqrt{p^3+1}}\ln \left|\frac{e^x+x-\sqrt{p^3+1}}{e^x+x+\sqrt{p^3+1}}\right|+C$ for $p>-1$

$I=\int \frac{(e^x+1)dx}{e^{2x}+x^2+2x{e}^x-p^3-1}$
$=\int \frac{(e^x+1)dx}{{(e^x+x)}^2-p^3-1}$
Put $e^x+x=t\Rightarrow (e^x+1)dx=dt$
$\therefore I=\int \frac{dt}{t^2-p^3-1}$

For $p=-1,$
$I=\int \frac{dt}{t^2}$
$=-\frac{1}{t}+C$
$=-\frac{1}{e^x+x}+C$


For $p<-1,$
$I=\int \frac{dt}{t^2+\underset{+ve}{\underset{}{(-p^3-1)}}}$
$=\frac{1}{\sqrt{-p^3-1}}{\tan }^{-1}\left(\frac{e^x+x}{\sqrt{-p^3-1}}\right)+C$


For $p>-1,$
$I=\int \frac{dt}{t^2-\underset{+ve}{\underset{}{(p^3+1)}}}$
$=\frac{1}{2\sqrt{p^3+1}}\ln \left|\frac{e^x+x-\sqrt{p^3+1}}{e^x+x+\sqrt{p^3+1}}\right|+C$