Question

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

Answer 1

Hint: Integration is the way of inverse process of differentiation. Instead of differentiating the function, you are given the distance of a function and asked to find its primitive i.e. original function. Such a process is called integration or anti-differentiation. Use integration by parts. Integration by parts is like the product rule of integration, in fact , it is derived. From the product rule it states that,
$\int{x{{e}^{x}}dx=x{{e}^{x}}-\int{{{e}^{x}}dx=x{{e}^{x}}-{{e}^{x}}+c}}$

Complete step by step answer:
As you know that,
The given equation is
${{x}^{3}}\sqrt{{{x}^{2}}+1}dx$
Here, substitute, $t=1+{{x}^{2}}$
$dt=2xdx$
Therefore after substituting the equation will be, apply integration on both sides,
$\int{{{x}^{3}}\sqrt{{{x}^{2}}+1}}dx=\dfrac{1}{2}\int{{{x}^{2}}\sqrt{1+{{x}^{2}}}\left( 2xdx \right)=\dfrac{1}{2}}$
$\int{\left( t-1 \right)\sqrt{t}}dt$
Now, solve the integral in $t$
$\dfrac{1}{2}\int{\left( t-1 \right)\sqrt{t}dt=\dfrac{1}{2}\int{t\sqrt{t}}dt-\dfrac{1}{2}\int{\sqrt{t}}dt}$
$\Rightarrow \dfrac{1}{2}\int{{{t}^{\dfrac{3}{2}}}dt-\dfrac{1}{2}\int{{{t}^{\dfrac{1}{2}}}dt}}$
$\Rightarrow\dfrac{1}{2}\dfrac{{{t}^{\dfrac{5}{2}}}}{\dfrac{5}{2}}-\dfrac{1}{2}\dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+c$
$\Rightarrow \dfrac{1}{2}\times \dfrac{2}{5}{{t}^{\dfrac{5}{2}}}-\dfrac{1}{2}\times \dfrac{2}{3}{{t}^{\dfrac{3}{2}}}+c$
Here, $\dfrac{5}{2}$ more to multiplication and get reciprocal and $\dfrac{3}{2}$ also converted into multiplication and get reciprocal and $8$ both $'2'$ in numerator and denominator of different fraction gets cancel.
$=\dfrac{2}{10}{{t}^{\dfrac{5}{2}}}-\dfrac{1}{2}\times \dfrac{2}{3}{{t}^{\dfrac{3}{2}}}+c$
Above, as to comes in $'2'$ table $'5'$ times therefore the fraction will gets converted into $\dfrac{1}{5}$
$\Rightarrow \dfrac{1}{5}{{t}^{\dfrac{5}{2}}}-\dfrac{1}{3}{{t}^{\dfrac{3}{2}}}+c$
$\Rightarrow \dfrac{1}{15}{{t}^{\dfrac{3}{2}}}\left( 3t-5 \right)+c$
Substitute again $1+{{x}^{2}}=t$ and $2xdx=dt$
$\int{{{x}^{3}}\sqrt{1+{{x}^{2}}}dx=\dfrac{1}{15}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{3}{2}}}\left( 3\left( 1+{{x}^{2}} \right)-5 \right)+c}$
$\Rightarrow \int{{{x}^{3}}\sqrt{1+{{x}^{2}}}dx=\dfrac{1}{15}\left( 1+{{x}^{2}} \right)\sqrt{1+{{x}^{2}}}\left( 3{{x}^{2}}-2 \right)+c}$

$\therefore$  The final answer is $\int{{{x}^{3}}\sqrt{1+{{x}^{2}}}dx=\dfrac{1}{15}}\left( 1+{{x}^{2}} \right)\sqrt{1+{{x}^{2}}}\left( 3{{x}^{2}}-2 \right)+c$

Additional Information:
Integration by parts is used when one has to integrate a product of two functions, integration of parts is useful.
If $f(x)=g(x).h(x)$
Then,
$\int{f(x)dx=g(x)\int{h(x)dx-\left( \int{\dfrac{d}{dx}g(x)\int{h(x)dx}} \right)}}dx$
This is called integration by parts the integral of the product of two functions may be given as, “First function into integral of the second minus integral of the derivative of the first into integral of the second” Which is nothing but. Integration by parts.
Now, one thing that must be noted is that the correct choice of first function and second function can either make or break a problem.
The correct choice can easily simplify and an incorrect one can put you in a lot of trouble.
For instance;
$\int{x.{{e}^{x}}dx}$ is the integral we need to evaluate.
If we use ${{e}^{x}}$ as the first function and $x$ as the second and integrate by parts.
$\int{x.{{e}^{x}}dx={{e}^{x}}\int{x.dx-\int{\left( {{e}^{x}}\int{x.dx} \right)dx}}}$
$=ex.\dfrac{{{x}^{2}}}{2}-\int{{{e}^{x}}.\dfrac{{{x}^{2}}}{2}}.dx+c$
If we apply integration by parts to be second term, we again get a term with a ${{x}^{3}}$ and so on.
This not only complicates the problem but spells disaster.
But if you had chosen $x$ to be the first and ${{e}^{x}}$ to be second. The integral would have been simply to evaluate.
$\int{x{{e}^{x}}dx=x\int{{{e}^{x}}dx-\int{\left( \dfrac{d}{dx}x\int{\left( {{e}^{x}}.dx \right)} \right)dx}}}$
$={{e}^{x}}.x-{{e}^{x}}+c$

Note: Apply integration by parts and replace the term. Of $'t'$ by $'x'$ as it gets easy to solve the problem and also consumes less time and again after getting the solution replace $'x'$ by $'t'$
Remember the rules of multiplication and division as if the denominator gets converted into a numerator it gets reciprocated.