Question

$\int \frac{x^3}{(x+1{)}^2}dx$ is equal to
(where $C$ is constant of integration)

• A. $\frac{x^2}{2}-2x+3\ln |x+1|+\frac{1}{x+1}+C$
• B. $\frac{x^2}{2}-x+2\ln |x+1|+\frac{1}{x+1}+C$
• C. $\frac{x^2}{2}+x+3\ln |x+1|+\frac{1}{x+1}+C$
• D. $\frac{x^2}{2}-2x+2\ln |x+1|+\frac{1}{x+1}+C$

Answer 1

The correct option is A $\frac{x^2}{2}-2x+3\ln |x+1|+\frac{1}{x+1}+C$

Adding and subtracting $1$ in the numerator, we get
$\int \frac{x^3}{(x+1{)}^2}dx=\int \frac{x^3+1-1}{(x+1{)}^2}dx$
$=\int \frac{x^3+1}{(x+1{)}^2}dx-\int \frac{1}{(x+1{)}^2}dx$
$=\int \frac{(x+1)(x^2-x+1)}{(x+1{)}^2}dx-\int \frac{1}{(x+1{)}^2}dx$
$=\int \frac{(x^2-x+1)}{(x+1)}dx-\int \frac{1}{(x+1{)}^2}dx$
$=\int (x-2+\frac{3}{x+1})dx-\int \frac{1}{(x+1{)}^2}dx$
$=\frac{x^2}{2}-2x+3\ln |x+1|+\frac{1}{x+1}+C$