Difficulty to find this integral

[DDarthVader]

Hello! I'm doing some calculus exercises and I'm having some difficulty to find the right answer.

Homework Statement

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) $u=\sqrt{5x+8}$

Homework Equations

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!

 

[DryRun]

(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.

 

[DDarthVader]

(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.

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

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

 

[Mentallic]

Hello! I'm doing some calculus exercises and I'm having some difficulty to find the right answer.

Homework Statement

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) $u=\sqrt{5x+8}$

Homework Equations

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?

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.

(b) Now $u=\sqrt{5x+8}$. Then $\frac{du}{dx}=\frac{1}{2\sqrt{5x+8}}$

You forgot about multiplying the expression by the derivative of 5x+8=5.

which means $dx=2u$

It should be $dx=2u\cdot du$

and now my integral is $2 \int du = u^2 = 5x+8$. Is this correct?
Thanks!

$$\int du= u$$

 

[DDarthVader]

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 $dx=2u\cdot du$


$$\int du= u$$

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

 

[Ray Vickson]

Hello! I'm doing some calculus exercises and I'm having some difficulty to find the right answer.

Homework Statement

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) $u=\sqrt{5x+8}$

Homework Equations

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!

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