Question

$\int \frac{\left(e^x+1\right)dx}{e^{2x}+x^2+2x{e}^x-p^3-1}$ is equal to
[Note: C is the constant of integration.]

• 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)

$I=\int \frac{\left(e^x+1\right)dx}{{\left(e^x+x\right)}^2-p^3-1}$
put $e^x+x=t\Rightarrow \left(e^x+1\right)dx=dt$
$I=\int \frac{dt}{t^2-p^3-1}$
(A) & (B) for $p<-1$

(C) for $p=-1$
$I=\frac{dt}{t^2}=\frac{-1}{t}+C$
(D) for $p>-1$
$I=\int \frac{dt}{t^2-\left(p^3+1\right)}=\frac{1}{2\sqrt{p^3+1}}\ln \left|\frac{t-\sqrt{p^3+1}}{t+\sqrt{p^3+1}}\right|+C$