1. ## Integration

I'm having a mental lapse...

I need a hint with respect to which technique I am supposed to be using to integrate the following:

$\int{\sqrt{1+\frac{9}{4}x}}dx$

2. Try u substitution.

3. Hint:

$\int \sqrt{ax+b} dx=(ax+b)^{\frac{3}{2}}\cdot \frac{2}{3a}$+C

4. At one point I had the following:

$\int{\frac{\sqrt{9x+4}}{2}}dx$

What makes a good u substitution in this case? It seems like picking a value of $u = \sqrt{1+\frac{9}{4}x}$ does not seem to help. I hope I am missing something blatently obvious.

5. Hint
If you have $\int f(x)\,dx=F(x)+C$ then $\int f(ax+b)\,dx=\dfrac{1}{a}F(ax+b)+C$

This can be proven by substituting $u=ax+b$ in your case $u=\dfrac{9}{4}x+1$

6. Yeah, I see now what I was doing wrong. Should have been:

$\int{1+(\frac{9}{4})x}dx$

let u = $1+(\frac{9}{4})x$

$\int{u^\frac{1}{2}\frac{4}{9}du}$

$\frac{4}{9}\int{u^\frac{1}{2}du}$

$\frac{4}{9}\frac{2}{3}u^\frac{3}{2}+C$

$\frac{8}{27}u^\frac{3}{2}+C$

$\frac{8}{27}(1+(\frac{9}{4})x)^\frac{3}{2}+C$

$\frac{8}{27}(\frac{(9x+4)^\frac{3}{2}}{4^\frac{3}{ 2}}))+C$

$\frac{8}{27}(\frac{(9x+4)^\frac{3}{2}}{8}))+C$

$\frac{(9x+4)^\frac{3}{2}}{8}+C$

...

And this was actually a definite integral such that $0\leq x \leq4$

So, I got a final answer of:

$\frac{80\sqrt{10}}{27}-\frac{8}{27}$

So...

$\frac{8(10\sqrt{10}-1)}{27}$

7. Your antiderivative is correct up until the last step. You need to cancel those 8's, right? What's left?

When you continue with plugging in limits, you appear to have corrected that error. Final answer is correct.

8. Whoops...

That should have been:

$\frac{(9x+4)^\frac{3}{2}}{27}+C$

9. Correct.