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

Hint: 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.