Question

Value of $\int e^x\frac{(x^2+3x+3)}{{(x+2)}^2}dx$ is

• A. $e^x\left[\frac{x-1}{x+2}\right]+c$
• B. $e^x\left[\frac{x+1}{x+2}\right]+c$
• C. $e^x\left[\frac{x+1}{x-2}\right]+c$
• D. None of these

Answer 1

The correct option is B $e^x\left[\frac{x+1}{x+2}\right]+c$

$\int e^x\left(\frac{x^2+3x+3}{{(x+2)}^2}\right)dx$

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

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

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

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

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

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

We have $\int e^x(f\left(x\right)+f^{\prime }\left(x\right))=e^{x}f\left(x\right)+c$

Using this we have $\int e^x(\frac{1}{x+2}-\frac{1}{{(x+2)}^2})dx=e^x\left(\frac{1}{x+2}\right)+c$

$\therefore \int e^x\left(\frac{x^2+3x+3}{{(x+2)}^2}\right)dx$

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

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

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

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

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