Question

Given, $\int{\dfrac{x}{{{\left( 1+x \right)}^{\dfrac{3}{2}}}}}dx=$
A) $2\left[ \dfrac{2+x}{\sqrt{1+x}} \right]+c$
B) $\left[ \dfrac{2+x}{\sqrt{1+x}} \right]+c$
C) $\dfrac{3}{2}\left[ \dfrac{x}{\sqrt{1-x}} \right]+c$
D) $\dfrac{3}{2}\left[ \dfrac{2+x}{\sqrt{1+x}} \right]+c$

Answer 1

In this problem we need to find an integral solution of the expression is given $\dfrac{x}{{{\left( 1+x \right)}^{\dfrac{3}{2}}}}$with respect to x, we have to consider the variable to simplify the problem that is $1+x=t$then after differentiating with respect to x we get $dx=dt$substitute this values in expression of integral and by further applying the basic integral formula like $\int{{{x}^{n}}}\,dx=\dfrac{{{x}^{n+1}}}{n+1}$we get the integral solution of the given expression.

Complete step by step answer:
In this problem, we need to find an integral solution of the given expression that is $\dfrac{x}{{{\left( 1+x \right)}^{\dfrac{3}{2}}}}$that is
$\int{\dfrac{x}{{{\left( 1+x \right)}^{\dfrac{3}{2}}}}}dx---(1)$
We have to consider $1+x=t$
After rearranging the term we get $x=t-1$
We have to differentiate with respect to x then we get:
$dx=dt$
After substituting the all the values of x in the given expression that is $\dfrac{x}{{{\left( 1+x \right)}^{\dfrac{3}{2}}}}$
After substituting we get:
$\dfrac{x}{{{\left( 1+x \right)}^{\dfrac{3}{2}}}}=\dfrac{t-1}{{{\left( t \right)}^{\dfrac{3}{2}}}}$
After splitting the terms we get:
$\dfrac{x}{{{\left( 1+x \right)}^{\dfrac{3}{2}}}}=\dfrac{t}{{{\left( t \right)}^{\dfrac{3}{2}}}}-\dfrac{1}{{{\left( t \right)}^{\dfrac{3}{2}}}}$
Simplifying the above equation we get:
$\dfrac{x}{{{\left( 1+x \right)}^{\dfrac{3}{2}}}}=t\times {{\left( t \right)}^{\dfrac{-3}{2}}}-{{\left( t \right)}^{\dfrac{-3}{2}}}$
By using the property of indices we have ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ substitute this property in thin above equation we get:
$\dfrac{x}{{{\left( 1+x \right)}^{\dfrac{3}{2}}}}={{\left( t \right)}^{1-\dfrac{3}{2}}}-{{\left( t \right)}^{\dfrac{-3}{2}}}$
Further simplifying this we get:
$\dfrac{x}{{{\left( 1+x \right)}^{\dfrac{3}{2}}}}={{\left( t \right)}^{\dfrac{-1}{2}}}-{{\left( t \right)}^{\dfrac{-3}{2}}}---(2)$
Substitute this value of equation (2) on equation (1) we get:
$\Rightarrow \int{\left[ {{\left( t \right)}^{\dfrac{-1}{2}}}-{{\left( t \right)}^{\dfrac{-3}{2}}} \right]\,}dt$
By applying the basic rule of integration that is $\int{{{x}^{n}}}\,dx=\dfrac{{{x}^{n+1}}}{n+1}$applying this formula on above equation we get:
$\Rightarrow \left[ \dfrac{{{\left( t \right)}^{\dfrac{-1}{2}+1}}}{\dfrac{-1}{2}+1}-\dfrac{{{\left( t \right)}^{\dfrac{-3}{2}+1}}}{\dfrac{-3}{2}+1} \right]\,$
By further simplifying this we get:
$\Rightarrow \left[ \dfrac{{{\left( t \right)}^{\dfrac{1}{2}}}}{\dfrac{1}{2}}-\dfrac{{{\left( t \right)}^{\dfrac{-1}{2}}}}{\dfrac{-1}{2}} \right]\,$
After simplifying further we get:
$\Rightarrow \left[ 2\sqrt{\left( t \right)}+\dfrac{2}{\sqrt{\left( t \right)}} \right]\,$
By cross multiplying this above equation we get:
$\Rightarrow \left[ \dfrac{2\left( t \right)+2}{\sqrt{\left( t \right)}} \right]\,$
By resubstituting the value of t that is $1+x=t$ in the above equation we get:
$\Rightarrow \left[ \dfrac{2\left( 1+x \right)+2}{\sqrt{1+x}} \right]\,$
By further solving this and adding the terms we get:
$\Rightarrow \left[ \dfrac{4+2x}{\sqrt{1+x}} \right]\,$
Take the 2 common things we get:
$\Rightarrow 2\left[ \dfrac{2+x}{\sqrt{1+x}} \right]\,$

So, the correct answer is “Option A”.

Note:
Always remember the important formulas for differentiation as well as integration. If not then the question becomes more difficult. In this problem we have assign the value of $\,t$ is $1+x=t$ but it is not mandatory you can assign any variable to make the problem easier and at last step don’t forget to substitute in the value of $\,t$ in the term of x. In this type of problem always first you simplify the problem and reduce it before integrating the expression otherwise integration becomes complicated.