How do you evaluate the indefinite integral #int(x + 34)sqrt(68x+x^2) dx#?

To evaluate the indefinite integral (\int (x + 34)\sqrt{68x+x^2} , dx), we can use trigonometric substitution. Let (x = a \sin(\theta)) or (x = a \cos(\theta)), where (a) is a constant. In this case, let's choose (x = \sqrt{68} \sin(\theta)). Then, (dx = \sqrt{68} \cos(\theta) d\theta).

Substituting (x = \sqrt{68} \sin(\theta)) and (dx = \sqrt{68} \cos(\theta) d\theta) into the integral, we have:

[ \begin{align*} \int (x + 34)\sqrt{68x+x^2} , dx &= \int (\sqrt{68} \sin(\theta) + 34) \sqrt{68(\sqrt{68} \sin(\theta)) + (\sqrt{68} \sin(\theta))^2} \sqrt{68} \cos(\theta) , d\theta \ &= \int (\sqrt{68} \sin(\theta) + 34)\sqrt{68(\sqrt{68} \sin(\theta)) + 68\sin^2(\theta)} , d\theta \ &= \int (\sqrt{68} \sin(\theta) + 34)\sqrt{68} \cos(\theta) , d\theta \ &= \int (\sqrt{68}\sin(\theta) + 34)(\sqrt{68}\cos(\theta)) , d\theta \ &= \int 68\sin(\theta)\cos(\theta) + 34\sqrt{68}\cos(\theta) , d\theta \ &= \int 34\sin(2\theta) + 34\sqrt{68}\cos(\theta) , d\theta. \end{align*} ]

Now, we can integrate term by term:

[ \begin{align*} \int 34\sin(2\theta) + 34\sqrt{68}\cos(\theta) , d\theta &= -\frac{34}{2}\cos(2\theta) + 34\sqrt{68}\sin(\theta) + C \ &= -17\cos(2\theta) + 34\sqrt{68}\sin(\theta) + C, \end{align*} ]

where (C) is the constant of integration. Finally, we can substitute back (x = \sqrt{68}\sin(\theta)) to express the solution in terms of (x):

[ \boxed{-17\cos\left(2\arcsin\left(\frac{x}{\sqrt{68}}\right)\right) + 34\sqrt{68}\sin\left(\arcsin\left(\frac{x}{\sqrt{68}}\right)\right) + C}. ]