Question

Integrate $\int \frac{(x^3+3x^2+3x+1)}{(x+1{)}^5}=dx$

• A. $-\frac{1}{x+1}+c$
• B. $\frac{1}{5}log(x+1)+c$
• C. $log(x+1)+c$
• D. $ta{n}^{-1}x+c$

Answer 1

The correct option is A $-\frac{1}{x+1}+c$

${\int }^{}\frac{x^3+3x^2+3x+1}{{(x+1)}^5}dx$

Simplify the above equation,

We get,$\frac{1}{(x+1{)}^2}$

Substitute $u=x+1\to dx=du$

$\begin{array}{c}{x}^2=(u-1{)}^2\\ x^3=(u-1{)}^3\end{array}$

$\begin{array}{c}={\int }^{}\frac{1}{u^2}du\\ {\int }^{}{u}^{n}du=\frac{u^{n+1}}{n+1}\end{array}$

$n=-2$

$=-\frac{1}{u}$

Since,$u=x+1$

$=-\frac{1}{x+1}+C$