Question

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

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

Answer 1

The correct option is A

$-\frac{1}{(x+1)}+c$

Explanation For The Correct Option:

Evaluating the integral:

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

$$\begin{gathered} =\int \frac{(x^3+1+3x(x+1\left)\right)}{{(x+1)}^5}dx \\ =\int \frac{{(x+1)}^3}{{(x+1)}^5}dx\left[\because {\left(a+b\right)}^3=a^3+b^3+3ab(a+b)\right] \\ =\int \frac{1}{{(x+1)}^2}dx \end{gathered}$$

Substituting $t=x+1$then $dx=dt$

$$\begin{gathered} \Rightarrow \int \frac{1}{t^2}dt \\ =\int t^{-2}dt \\ =-\frac{1}{t^{}}+c[c=\text{constant}] \\ =-\frac{1}{(x+1)}+c\left[\text{substituting back}t=(x+1)\right] \end{gathered}$$

Hence, option (A) is the correct answer.