Question

$\int e^x\frac{x-1}{{\left(x+1\right)}^3}\text{dx}\text{equals}$

• A. $-\frac{e^x}{x+1}+c$
• B. $\frac{e^x}{x+1}+c$
• C. $\frac{e^x}{{(x+1)}^2}+c$
• D. $-\frac{e^x}{{(x+1)}^2}$

Answer 1

The correct option is C $\frac{e^x}{{(x+1)}^2}+c$

Solution:-

$I=\int e^x.\frac{x-1}{(x+1{)}^3}dx$

$=\int e^x.\frac{x+1-2}{(x+1{)}^3}dx$

$=\int e^x.\{\frac{x+1}{(x+1{)}^3}-\frac{2}{(x+1{)}^3}\}dx$

$=\int \{\frac{e^x}{(x+1{)}^2}-\frac{2e^x}{(x+1{)}^3}\}dx$

Let $\frac{e^x}{(x+1{)}^2}=z$

$\therefore \left\{\frac{e^x(x+1{)}^2-x^2.2(2x+1)}{(x+1{)}^4}\right\}dx=dz$

$\{\frac{e^x}{(x+1{)}^2}-\frac{2e^x}{(x+1{)}^3}\}dx=dz$

$\therefore I=\int dz=z+c$

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