Question

Evaluate the following integral
\[\int{\left( x-3 \right)\sqrt{{{x}^{2}}+3x-18}dx}\]

Answer 1

Hint: Whenever we have to evaluate integration that involves terms under square root, we always try to assume terms under square root a new variable, say $t$ . Now, at this step we will try to make the whole integral in terms of $t$ . For that we try to have the differentiation of ${{x}^{2}}+3x-18$ outside the square root so that we can easily convert the integral with respect to $t$.

Complete step by step answer:
Consider the given integral,
\[\int{\left( x-3 \right)\sqrt{{{x}^{2}}+3x-18}dx}\]
The given integrand seems complicated, we cannot directly integrate it. We have to simplify it, so that we can integrate it.
The integrand involves a big term under square root, so to simplify it, first we will assume the big term a new variable, say $t$ ,
Let, ${{x}^{2}}+3x-18=t$
Now, differentiating with respect to $''x''$ on both the sides, we get,
$\left( 2x+3 \right)dx=dt$
Now, we have to we’ll try to make $2x+3$ outside the integration so that, we can easily convert the integral in terms of $t$ ,
Since, we have to twice of $x$ , so we will first multiply and the integral the integral by 2 and get,
\[\dfrac{1}{2}\int{\left( 2x-6 \right)\sqrt{{{x}^{2}}+3x-18}dx}\]
Now, we also need $+3$ after $2x$ , so we will subtract and add $3$ to $2x-6$ , and get,
\[\dfrac{1}{2}\int{\left( 2x+3-9 \right)\sqrt{{{x}^{2}}+3x-18}dx}\]
Now, we will separate the terms,
\[\dfrac{1}{2}\int{\left( 2x+3 \right)\sqrt{{{x}^{2}}+3x-18}}dx-\dfrac{9}{2}\int{\sqrt{{{x}^{2}}+3x-18}dx}\]
Now, convert the first integral in terms of $t$ and leave the second integral as it is, and get,
\[\dfrac{1}{2}\int{\sqrt{t}dt}-\dfrac{9}{2}\int{\sqrt{{{x}^{2}}+3x-18}dx}\]
Integrate the first integral and write the second integral as(using the completing the square method),
\[\dfrac{1}{3}{{t}^{\dfrac{3}{2}}}-\dfrac{9}{2}\int{\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{9}{2} \right)}^{2}}}dx}\]
Now, put the value of t in the first term and get,
\[\dfrac{1}{3}{{\left( {{x}^{2}}+3x-18 \right)}^{\dfrac{3}{2}}}-\dfrac{9}{2}\int{\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{9}{2} \right)}^{2}}}dx}\]
We know the formula of integration that,
$\int{\sqrt{{{x}^{2}}-{{a}^{2}}}dx}=\dfrac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\dfrac{{{a}^{2}}}{2}\ln \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C$
Applying this formula in the last step, we get
\[\begin{align}
  & \dfrac{1}{3}{{\left( {{x}^{2}}+3x-18 \right)}^{\dfrac{3}{2}}}-\dfrac{9}{2}\left[ \dfrac{\left( x+\dfrac{3}{2} \right)}{2}\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{9}{2} \right)}^{2}}}-\dfrac{{{\left( \dfrac{9}{2} \right)}^{2}}}{2}\ln \left| \left( x+\dfrac{3}{2} \right)+\sqrt{{{\left( x+\dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{9}{2} \right)}^{2}}} \right| \right]+C \\
 & =\dfrac{1}{3}{{\left( {{x}^{2}}+3x-18 \right)}^{\dfrac{3}{2}}}-\dfrac{9}{8}\left[ \left( 2x+3 \right)\sqrt{{{x}^{2}}+3x-18}-\dfrac{81}{2}\ln \left| \left( x+\dfrac{3}{2} \right)+\sqrt{{{x}^{2}}+3x-18} \right| \right]+C \\
\end{align}\]

Hence,
\[\int{\left( x-3 \right)\sqrt{{{x}^{2}}+3x-18}dx}=\dfrac{1}{3}{{\left( {{x}^{2}}+3x-18 \right)}^{\dfrac{3}{2}}}-\dfrac{9}{8}\left[ \left( 2x+3 \right)\sqrt{{{x}^{2}}+3x-18}-\dfrac{81}{2}\ln \left| \left( x+\dfrac{3}{2} \right)+\sqrt{{{x}^{2}}+3x-18} \right| \right]+C\]

Note: Note that whenever we see an integral with a big term under fractional powers, we always try to assume that term a new variable , because integration under fractional power is a complicated task. We do not have any direct integration formula. So, whenever you see a big term under fractional power, assume that term a new variable and as same as this question try to have differentiation of big term outside the fractional power and then integrate in terms of the new variable.