Question

Integrate the integral below:

Answer 1

\[\int\limits\limits_{}^{}\frac{ xdx }{ (5-2x)^{\frac{ 3 }{2 }} }\]
Looking at it, I think the only way to solve the integral above is by using substitution.
So I let
\[t= (5-2x)^{\frac{ 3 }{ 2 }}\]
from this I got the value of
\[x= \frac{ 5-t ^{\frac{ 2 }{ 3 }} }{ 2 }\]
which is
\[dx= -\frac{ dt }{ 3t ^{\frac{ 1 }{ 3 }} }\]
^I am not very sure if I got the derivative right
Then I need to substitute the values to the original integral, which will be equal to
\[\int\limits\limits_{}^{}\frac{ \frac{ 5dt-t ^{\frac{ 2 }{ 3 }}dt }{ 6t ^{\frac{ 1 }{ 3 }} } }{ t }\]
Then after this I don't know what to do anymore, I'm not even sure if I am on the right track. Help?

Answer 2

Try \[t = (5-2x)^{1/2}\]

Answer 3

why did you use 1/2 instead of 3/2?

Answer 4

Let me calculate in my rough notebook.

Answer 5

ok, I'll wait :)
But what is the reason on why you advised me to use 1/2?

Answer 6

How about u=5-2x
du=-2dx
-du/2=dx
\[\huge -\frac{1}{2} \int\limits \frac{5-u}{2 u^{\frac{3}{2}}} \, du\]

Answer 7

no bro! I think I got it.

Answer 8

\[\frac{dt}{dx} = \frac{-1}{(5-2x)^{1/2}} \]

Answer 9

Split and integrate separately
\[-\frac{1}{2} \int\limits \frac{5}{2 u^{\frac{3}{2}}}-\frac{1}{2 \sqrt{u}} \, du\]

Answer 10

\[x = \frac{5-t^2}{2}\]

Answer 11

integral becomes
\[\int\limits_{}^{} \frac{(5-t^2) dt}{2t^2}\]

Answer 12

Go ahead @taiga_aisaka

Answer 13

Did you still used \[t= (5-2x)^{\frac{ 1 }{ 2 }}?\]

Answer 14

Yes I did this using that substitution !

Answer 15

Ok. Thank you very much.

Answer 16

First tell that did you get the right answer ? :P lol

Answer 17

Not yet, I'm currently solving it.