Question

Integrate the function: $x\sqrt{x+2}$

Answer 1

Hint: We have to find the value of the given indefinite integral. We will use the substitution method to integrate the given function. After we integrate the function with the substituted variable, we will resubstitute the value of the new variable to get the required answer.

Complete step by step answer:
We have to integrate the following, $\int{x\sqrt{x+2}dx}$.
Let $t=\sqrt{x+2}$. Now, we will have to find the value of $x$ and $dx$ in terms of $t$. As $t=\sqrt{x+2}$, we have the following equation after squaring both sides, \[{{t}^{2}}=x+2\]. Therefore, $x={{t}^{2}}-2$.
To find the value of $dx$ in terms of $t$, we will differentiate the equation $t=\sqrt{x+2}$. So we get the following,
\[\begin{align}
  & dt=\dfrac{1}{2\sqrt{x+2}}dx \\
 & dt=\dfrac{1}{2t}dx \\
\end{align}\]
Hence, $dx=2tdt$.
Now we have $x={{t}^{2}}-2$ ; $\sqrt{x+2}=t$ and $dx=2tdt$.We will substitute these values in the integral as follows,\[\int{({{t}^{2}}-2)\cdot t\cdot 2tdt}\] . Let the integral be $I$. We will now evaluate this integral in the following manner,
\[\begin{align}
  & I=\int{({{t}^{2}}-2)\cdot t\cdot 2tdt} \\
 & =\int{({{t}^{2}}-2)\cdot 2{{t}^{2}}dt} \\
 & =\int{(2{{t}^{4}}-4{{t}^{2}})dt} \\
 & =\int{2{{t}^{4}}dt}-\int{4{{t}^{2}}dt}
\end{align}\]
Now, we know that $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$, where $c$ is a constant. Using this formula, the value of the integral will be,
$I=\dfrac{2{{t}^{5}}}{5}-\dfrac{4{{t}^{3}}}{3}+C$ , where $C$ is a constant.
Now we will resubstitute the value of $t=\sqrt{x+2}$, we get the following equation,
\[I=\dfrac{2{{\left( \sqrt{x+2} \right)}^{5}}}{5}-\dfrac{4{{\left( \sqrt{x+2} \right)}^{3}}}{3}+C\]

So, the value of the integral is \[\int{x\sqrt{x+2}dx}=\dfrac{2{{(x+2)}^{{}^{5}/{}_{2}}}}{5}-\dfrac{4{{(x+2)}^{{}^{3}/{}_{2}}}}{3}+C\]

Note: The method used for evaluating the integral depends on the function which is to be integrated. In this question, we used the method of substitution. The other methods to evaluate integrals are by parts and by partial fractions. The integral in this question can also be evaluated by parts, but the calculations become complicated. Hence, integration by substitution was the best choice for this question.