Difficulty to find this integral

  • #1
Hello! I'm doing some calculus exercises and I'm having some difficulty to find the right answer.

Homework Statement


The integral is [itex]\int \frac{1}{\sqrt{5x+8}}dx[/itex] and the exercise tell me to solve this by using (a) [itex]u=5x+8[/itex] and (b) [itex]u=\sqrt{5x+8}[/itex]

Homework Equations





The Attempt at a Solution


This is my attempt:
(a) Well, [itex]u=5x+8[/itex] then [itex]\frac{du}{dx}=5[/itex] which means [itex]dx=\frac{du}{5}[/itex]. Then my integral is now [itex]\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}[/itex] Is this correct?

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

Answers and Replies

  • #2
DryRun
Gold Member
838
4
(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
(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 [itex]\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}}[/itex]
 
  • #4
Mentallic
Homework Helper
3,798
94
Hello! I'm doing some calculus exercises and I'm having some difficulty to find the right answer.

Homework Statement


The integral is [itex]\int \frac{1}{\sqrt{5x+8}}dx[/itex] and the exercise tell me to solve this by using (a) [itex]u=5x+8[/itex] and (b) [itex]u=\sqrt{5x+8}[/itex]

Homework Equations





The Attempt at a Solution


This is my attempt:

(a) Well, [itex]u=5x+8[/itex] then [itex]\frac{du}{dx}=5[/itex] which means [itex]dx=\frac{du}{5}[/itex]. Then my integral is now [itex]\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}[/itex] Is this correct?
No. While [tex]\int \frac{dx}{x}=\ln(x)[/tex] 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
[tex]\int x^n dx=\frac{x^{n+1}}{n+1}[/tex] where n=-1/2 in this case.

(b) Now [itex]u=\sqrt{5x+8}[/itex]. Then [itex]\frac{du}{dx}=\frac{1}{2\sqrt{5x+8}}[/itex]
You forgot about multiplying the expression by the derivative of 5x+8=5.
which means [itex]dx=2u[/itex]
It should be [itex]dx=2u\cdot du[/itex]

and now my integral is [itex]2 \int du = u^2 = 5x+8[/itex]. Is this correct?
Thanks!!
[tex]\int du= u[/tex]
 
  • #5
No. While [tex]\int \frac{dx}{x}=\ln(x)[/tex] 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
[tex]\int x^n dx=\frac{x^{n+1}}{n+1}[/tex] where n=-1/2 in this case.


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

It should be [itex]dx=2u\cdot du[/itex]


[tex]\int du= u[/tex]

Wow! Lots of stupid mistakes. Now I see what I did wrong! Thank you guys very much!
 
  • #6
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
Hello! I'm doing some calculus exercises and I'm having some difficulty to find the right answer.

Homework Statement


The integral is [itex]\int \frac{1}{\sqrt{5x+8}}dx[/itex] and the exercise tell me to solve this by using (a) [itex]u=5x+8[/itex] and (b) [itex]u=\sqrt{5x+8}[/itex]

Homework Equations





The Attempt at a Solution


This is my attempt:
(a) Well, [itex]u=5x+8[/itex] then [itex]\frac{du}{dx}=5[/itex] which means [itex]dx=\frac{du}{5}[/itex]. Then my integral is now [itex]\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}[/itex] Is this correct?

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

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

Anyway, your first approach, giving [tex]\frac{1}{5}\ln \sqrt{5x+8}[/tex] is wrong. The integral
[tex] \int \frac{du}{\sqrt{u}}[/tex] is of the form
[tex] \int u^n \, du, [/tex]
with [itex] n \neq -1,[/itex] 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
 

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