Question

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

• A. $\frac{e^x}{x+1}=C$
• B. $\frac{-e^x}{x-1}=C$
• C. $e^x\left(\frac{x-1}{x+1}\right)=C$
• D. $e^x\frac{(x+1)}{x-1}+C$

Answer 1

Answer: (C)

$e^x\left(\frac{x-1}{x+1}\right)=C$

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

Let $f\left(x\right)=\frac{x-1}{x+1}$

then $f^{\prime }\left(x\right)=\frac{x+1-x+1}{(x+1{)}^2}=\frac{2}{(x+1{)}^2}$

$f\left(x\right)+f^{\prime }\left(x\right)=\frac{x-1}{x+1}+\frac{2}{(x+1{)}^2}=\frac{x^2+1}{(x+1{)}^2}$

As, $\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx=e^{x}f\left(x\right)$

So,I $=\left(\frac{x-1}{x+1}\right)$