Question

Integral of (x-3)root(x^2+3x-18)
Can anyone help,how to do this type of problems?

Answer 1

\[\int\limits_{ }^{ }(x-3)\sqrt{x^2+3x-18}~~dx \]

Answer 2

\[= (\frac{1}{3}x^3-3x)\frac{2}{3}(x^2+3x-18)^{3/2} -\int\limits_{ }^{ }\sqrt{x^2+3x-18}~~dx\](integrating by parts)

Answer 3

\[= (\frac{2}{9}x^3-3x)(x^2+3x-18)^{3/2} -\int\limits_{ }^{ }\sqrt{x^2+3x-18}~~dx\]simplified a but.

Answer 4

then u substitution for the second integral.
\[\int\limits_{ }^{ }\sqrt{x^2+3x-18}~~dx\]Completing the square inside the root,\[\int\limits_{ }^{ }\sqrt{x^2+3x+2.25-2.25-18}~~dx\]\[\int\limits_{ }^{ }\sqrt{x^2+3x+2.25-20.25}~~dx\]\[\int\limits_{ }^{ }\sqrt{(x+1.5)^2-20.25}~~dx\]\[u=x+1.5~~~~~~du=dx\]\[\int\limits_{ }^{ }\sqrt{(u)^2-20.25}~~du\]

Answer 5

So it is a bit better now.

Answer 6

Another approach that (mostly) avoids IBP:
\[\int (x-3)\sqrt{x^2+3x-18}~dx\]
Complete the square:
\[\begin{align*}
x^2+3x-18&=x^2+3x+\frac{9}{4}-\frac{81}{4}\\\\
&=\left(x+\frac{3}{2}\right)^2-\frac{81}{4}
\end{align*}\]
You may also write
\[\begin{align*}
x-3&=x+\frac{3}{2}-\frac{9}{2}
\end{align*}\]
so the integral is equivalent to
\[\int \left(x+\frac{3}{2}\right)\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{81}{4}}~dx-\frac{9}{2}\int\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{81}{4}}~dx\]
In the first integral, you can substitute \(u=\left(x+\dfrac{3}{2}\right)^2-\dfrac{81}{4}\) so that \(\dfrac{du}{2}=\left(x+\dfrac{3}{2}\right)~du\). With the second integral, you can use \(\dfrac{9}{2}\sec t=x+\dfrac{3}{2}\), giving \(\dfrac{9}{2}\sec t\tan t~dt=dx\).
\[\frac{1}{2}\int \sqrt{u}~du-\left(\frac{9}{2}\right)^2\int\sqrt{\left(\frac{9}{2}\sec t\right)^2-\frac{81}{4}}~\sec t\tan t~dt\]
\[\frac{1}{2}\int \sqrt{u}~du-\left(\frac{9}{2}\right)^3\int\sqrt{\sec^2t-1}~\sec t\tan t~dt\]
\[\frac{1}{2}\int \sqrt{u}~du-\left(\frac{9}{2}\right)^3\int\sec t\tan^2 t~dt\]
\[\frac{1}{2}\int \sqrt{u}~du-\left(\frac{9}{2}\right)^3\int(\sec^3t-\sec t)~dt\]
You can use IBP to determine the integral of secant cubed, or make use of the power-reduction formula.