What is the most efficient way to integrate $(x-3)\sqrt{x^2+3x-18}$?

Question

I can do the problem, but it is becoming so big,that I do not feel to do it anymore. Can anyone give the shortest method for this problem?

$$\int (x-3)\sqrt{x^2+3x-18}\,dx $$

Answer 1

As $$x^2+3x-18=\frac{4x^2+12x-72}4=\frac{(2x+3)^2-9^2}4$$

Using Trigonometric substitution, set $2x+3=9\sec\theta$

So, we have $x-3=\dfrac92(\sec\theta-1)\implies dx=\dfrac92\sec\theta\tan\theta\ d\theta$

$x^2+3x-18=\dfrac{9\tan^2\theta}4\implies\sqrt{x^2+3x-18}=\dfrac32|\tan\theta|$

$$I=\int(x-3)\sqrt{x^2+3x-18}\ dx=\dfrac32|\tan\theta|\dfrac92(\sec\theta-1)\dfrac92\sec\theta\tan\theta\ d\theta$$

If $\tan\theta\ge0,$ $$I=\frac{3^5}8\int (\sec\theta-1)\sec\theta\tan^2\theta\ d\theta$$

$$\frac8{3^5}I=\int(\sec^2\theta\tan^2\theta)\ d\theta-\int\sec\theta\tan^2\theta\ d\theta$$

$$=\int d(\tan^2\theta)-\int\sec\theta(\sec^2\theta-1)\ d\theta$$

$$=\frac{\tan^3\theta}3-\int\sec^3\theta\ d\theta+\int\sec\theta\ d\theta$$

Now use Integral of the secant function and Integral of secant cubed

Answer 2

Hint: substitute $x=u+3$. Then,

$$\int (x-3)\sqrt{x^2+3x-18}\,dx =\int u\sqrt{u(u+9)}\,du$$

Answer 3

As $\dfrac{d(x^2+3x-18)}{dx}=2x+3$

$$I=\int(x-3)\sqrt{x^2+3x-18}\ dx,$$

$$2I=\int(2x-6)\sqrt{x^2+3x-18}\ dx$$ $$=\int(2x+3)\sqrt{x^2+3x-18}\ dx-9\sqrt{x^2+3x-18}\ dx$$

For the second integral, $x^2+3x-18=\left(x+\dfrac32\right)^2-\left(\dfrac92\right)^2,$

set $x+\dfrac32=u$ and use this