Question

How do you integrate \[{{x}^{3}}\sqrt{{{x}^{2}}+1}dx\] ?

Answer 1

Hint: We can solve this question by using Integration by substitution method. We can replace \[{{x}^{2}}+1\] as u and its derivative as du. From this we can solve our new function formed after substitution using basic integration techniques until we arrive at the solution.

Complete step by step answer:
Before going to the solution let us know about the Integration by substitution method.
The substitution method is used when an integral contains some function and its derivative. In this case, we can set u equal to the function and rewrite the integral in terms of the new variable u. This makes the integral easier to solve.
Substitution Rule:
\[\int{f\left( g\left( x\pi \right) \right)g'\left( x \right)dx=\int{f\left( u \right)du}}\] where, \[u=g\left( x \right)\]
After solving the function we have to back substitute x in place of u to get the answer of our function in original variables.
Given function is \[\int{{{x}^{3}}\sqrt{{{x}^{2}}+1}dx}\]
Now take \[{{x}^{2}}+1\] as u then
\[u={{x}^{2}}+1\]
\[du=2xdx\]
But in the function we don’t have 2dx so we divide and multiply the equation with 2
\[\Rightarrow \dfrac{1}{2}\int{2{{x}^{3}}\sqrt{{{x}^{2}}+1}dx}\]
Now rewriting \[2x\] in the starting with \[dx\] so that we get \[2xdx\]
 \[\Rightarrow \dfrac{1}{2}\int{{{x}^{2}}\sqrt{{{x}^{2}}+1}2xdx}\]
Now substitute \[u={{x}^{2}}+1\] and \[du=2xdx\]
Then the equation will become
\[\Rightarrow \dfrac{1}{2}\int{{{x}^{2}}\sqrt{u}du}\]
Now we can write \[{{x}^{2}}\] as \[u-1\] from \[u={{x}^{2}}+1\]
Then the equation will be
\[\Rightarrow \dfrac{1}{2}\int{\left( u-1 \right)\sqrt{u}du}\]
From this we can rewrite the equation as
\[\Rightarrow \dfrac{1}{2}\int{u\sqrt{u}du}-\dfrac{1}{2}\int{\sqrt{u}du}\]
\[\Rightarrow \dfrac{1}{2}\int{{{u}^{\dfrac{3}{2}}}du-\dfrac{1}{2}\int{{{u}^{\dfrac{1}{2}}}du}}\]
Now integrating the function using power rule
Power rule:
\[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}\]
Here we will substitute \[\dfrac{3}{2}\] and \[\dfrac{1}{2}\] in place of n in power rule
So the equation will become
\[\Rightarrow \dfrac{1}{2}\dfrac{{{u}^{\dfrac{5}{2}}}}{\dfrac{5}{2}}-\dfrac{1}{2}\dfrac{{{u}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+c\]
By simplifying the above equation we will get
\[\Rightarrow \dfrac{1}{5}{{u}^{\dfrac{5}{2}}}-\dfrac{1}{3}{{u}^{\dfrac{3}{2}}}+c\]
Now subtracting the same variable containing terms we will get
\[\Rightarrow \dfrac{1}{15}{{u}^{\dfrac{3}{2}}}\left( 3u-5 \right)+c\]
Now we have arrived at the simplest form of the equation.
But to get the answer for original function we have to back substitute
\[u={{x}^{2}}+1\]
Then we will get
\[\Rightarrow \dfrac{1}{15}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{3}{2}}}\left( 3\left( 1+{{x}^{2}} \right)-5 \right)+c\]
By further simplifying it we will get
\[\Rightarrow \dfrac{1}{15}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{3}{2}}}\left( 3{{x}^{2}}+3-5 \right)+c\]
\[\Rightarrow \dfrac{1}{15}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{3}{2}}}\left( 3{{x}^{2}}-2 \right)+c\]
\[\Rightarrow \dfrac{1}{15}\left( 1+{{x}^{2}} \right){{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}\left( 3{{x}^{2}}-2 \right)+c\]
After the simplifications applied we will get
\[\dfrac{1}{15}\left( 1+{{x}^{2}} \right)\sqrt{\left( 1+{{x}^{2}} \right)}\left( 3{{x}^{2}}-2 \right)+c\]
So we can write \[{{x}^{3}}\sqrt{{{x}^{2}}+1}dx=\dfrac{1}{15}\left( 1+{{x}^{2}} \right)\sqrt{\left( 1+{{x}^{2}} \right)}\left( 3{{x}^{2}}-2 \right)+c\].

Note:
Students must be careful while using Integration by substitution method as if we don’t do the back substitute part we won’t get the required solution. We will also get slight differences in the answer because there are different ways to simplify the equation after back substitution.