Question

$\int e^x\frac{(x^2-3)}{(x+3{)}^2}dx$ is equal to

• A. $e^x\left(\frac{x}{x+3}\right)+c$
• B. $e^x(2-\frac{6}{x+3})+c$
• C. $e^x(1-\frac{6}{x+3})+c$
• D. $e^x\left(\frac{3}{x+3}\right)+c$

Answer 1

The correct option is C $e^x(1-\frac{6}{x+3})+c$

$I=\int e^x\frac{(x^2-3)}{(x+3{)}^2}dx=\int e^x\frac{(x^2-9+6)}{(x+3{)}^2}dx=\int e^x(\frac{x-3}{x+3}+\frac{6}{(x+3{)}^2})dx\underset{f\left(x\right)}{\downarrow }\underset{f^{\prime }\left(x\right)}{\downarrow }=\int e^x\left(f\right(x)+f^{{}^{\prime }}(x\left)\right)dx=e^{x}f\left(x\right)+c=e^x\frac{(x-3)}{(x+3)}+c=e^x(1-\frac{6}{x+3})+c$