I'm uncertain about something. I thought this was a simple problem where you simply bring up the denominator and apply the general power rule, but that doesn't seem to be right... Can anyone point me in the right direction?
I'm uncertain about something. I thought this was a simple problem where you simply bring up the denominator and apply the general power rule, but that doesn't seem to be right... Can anyone point me in the right direction?
Have you been taught the substitution technique.
Otherwise you will just have to use the following result which I'm sure is somewhere in your class notes or textbook:
$\displaystyle \int (ax + b)^n \, dx = \frac{1}{a(n+1)} (ax + b)^{n + 1}$ provided $\displaystyle n \neq -1$ (and I have ignored the arbitrary constant since it's not needed for a definite integral).
For the first problem: Let $\displaystyle u = \sqrt{4x+1}$, then $\displaystyle \dfrac{du}{dx} = \dfrac{2}{\sqrt{(4x+1)}} \Rightarrow dx = \dfrac{\sqrt{4x+1}}{2}\;{du}$ so that $\displaystyle \int_{x=6}^{x=20}\dfrac{dx}{\sqrt{4x+1}} = \int_{u=5}^{u=9}\left(\dfrac{1}{u}\right)\left(\df rac{u}{2}\right)\;{du} $ and finish off.
For the second problem: Let again $\displaystyle u = \sqrt{4x+1}$, then $\displaystyle \dfrac{du}{dx} = \dfrac{2}{\sqrt{(4x+1)}} \Rightarrow dx = \dfrac{\sqrt{4x+1}}{2}\;{du}$ so that $\displaystyle \dfrac{1}{4}\int_{x=6}^{x=20}\dfrac{4}{\sqrt{4x+1} }\;{dx} = \dfrac{1}{4}\int_{u=5}^{u=9}\left(\dfrac{4}{u}\rig ht)\left(\dfrac{u}{2}\right)\;{du} $ and finish off.