Question

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

• A. $2\left[\frac{(2+x)}{\sqrt{(1+x)}}\right]+c$
• B. $\left[\frac{(2+x)}{\sqrt{(1+x)}}\right]+c$
• C. $\frac{3}{2}\left[\frac{x}{\sqrt{(1-x)}}\right]+c$
• D. $\frac{3}{2}\left[\frac{(2+x)}{\sqrt{(1+x)}}\right]+c$

Answer 1

The correct option is A

$2\left[\frac{(2+x)}{\sqrt{(1+x)}}\right]+c$

Explanation For The Correct Option:

Evaluating the integral

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

Substituting $\left(1+x\right)=t^2$

Differentiating it w.r.t $x$ then $dx=2tdt$

$$\begin{gathered} \int \frac{x}{{\left(1+x\right)}^{{\displaystyle \frac{3}{2}}}}=\int \frac{\left(t^2-1\right)(2tdt)}{{\left(t^2\right)}^{{\displaystyle \frac{3}{2}}}} \\ =2\int \frac{t\left(t^2-1\right)}{t^3}dt \\ =2\int \left(1-\frac{1}{t^2}\right)dt \\ =2\int \left(1-t^{-2}\right)dt \\ =2\left(t+t^{-1}\right)\left[\because \int x^{n}dx=\frac{x^{n+1}}{n+1}+c\right] \\ =2\left(t+\frac{1}{t}\right) \\ =2\left(\frac{t^2+1}{t}\right)+c[c=\text{constant}] \\ =2\frac{x+2}{\sqrt{1+x}}+c[\text{substituting back}\left(1+x\right)=t^2] \end{gathered}$$

Hence, the correct answer is option (A).