# Homework Help: Difficulty to find this integral

1. Jun 14, 2012

Hello! I'm doing some calculus exercises and I'm having some difficulty to find the right answer.
1. The problem statement, all variables and given/known data
The integral is $\int \frac{1}{\sqrt{5x+8}}dx$ and the exercise tell me to solve this by using (a) $u=5x+8$ and (b) $u=\sqrt{5x+8}$

2. Relevant equations

3. The attempt at a solution
This is my attempt:
(a) Well, $u=5x+8$ then $\frac{du}{dx}=5$ which means $dx=\frac{du}{5}$. Then my integral is now $\int \frac{du}{5 \sqrt{u}} = \frac{1}{5}ln_{\sqrt{u}} = \frac{1}{5}ln_{\sqrt{5x+8}} = \frac{1}{5}ln_{5x}ln_{8}$ Is this correct?

(b) Now $u=\sqrt{5x+8}$. Then $\frac{du}{dx}=\frac{1}{2\sqrt{5x+8}}$ which means $dx=2u$ and now my integral is $2 \int du = u^2 = 5x+8$. Is this correct?
Thanks!!

2. Jun 14, 2012

### sharks

(a) $\int \frac{du}{5 \sqrt{u}}=\frac{1}{5}\int \frac{du}{ \sqrt{u}}=\frac{2\sqrt u}{5}+C$
I get the final result from the general formula (which is very helpful if you remember it): $\int \frac{f'(x)}{\sqrt{f(x)}}\,dx=2\sqrt{f(x)}+C$

(b) $\frac{du}{dx}=\frac{1}{2\sqrt{5x+8}}$ is wrong.

3. Jun 14, 2012

Oh, I don't know this formula. Thanks!

About the (b), why is it wrong? This is what I did $\frac{du}{dx}=\frac{d(\sqrt{5x+8})}{dx}\; \Rightarrow \frac{du}{dx}=((5x+8)^{1/2})' = \frac{1}{2}(5x+8)^{\frac{1}{2}-1}= \frac{1}{2}(5x+8)^{-\frac{1}{2}} = \frac{1}{2}\frac{1}{\sqrt{5x+8}}=\frac{1}{2\sqrt{5x+8}}$

4. Jun 14, 2012

### Mentallic

No. While $$\int \frac{dx}{x}=\ln(x)$$ you need to keep in mind that the integral only becomes a logarithm when the denominator is a linear equation in x (so of the form ax+b). We don't have that, instead you need to use the rule
$$\int x^n dx=\frac{x^{n+1}}{n+1}$$ where n=-1/2 in this case.

You forgot about multiplying the expression by the derivative of 5x+8=5.
It should be $dx=2u\cdot du$

$$\int du= u$$

5. Jun 14, 2012

Wow! Lots of stupid mistakes. Now I see what I did wrong! Thank you guys very much!

6. Jun 14, 2012

### Ray Vickson

NO, NO, NO! Why would you ever think that $\ln(5x+8) = \ln(5x) \ln(8)$? For that matter, what do you mean by the notation $\ln_{\sqrt{u}}$? Did you mean to write
$\ln \sqrt{u}$ or $\ln(\sqrt{u})$?

Anyway, your first approach, giving $$\frac{1}{5}\ln \sqrt{5x+8}$$ is wrong. The integral
$$\int \frac{du}{\sqrt{u}}$$ is of the form
$$\int u^n \, du,$$
with $n \neq -1,$ so can be integrated without involving logarithms.

Your second approach is wrong: you have the wrong "du".

BTW: to get proper typesetting of math names like log, ln, sin, cos, exp, etc. in LaTeX, you need to enter them as commands, like this: \log, \ln, \sin, \cos, \exp, etc.

RGV