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# Integration by substitution. watch

1. So I really cant get very far with this. I've been ok with substitution but with this one having a fraction in it, I'm struggling.

$\int\frac{5x}{5x^{2}+7} dx$
2. (Original post by Banny Dyrne)
So I really cant get very far with this. I've been ok with substitution but with this one having a fraction in it, I'm struggling.

$\int\frac{5x}{5x^{2}+7} dx$
Have you been given a substitution to use or do you need to work one out?
3. Numerator is half the differential of the denominator so the integral is 1/2log(5x^2 + 7)
4. Yeah the substitution is $u = 5x^{2}+7$
5. (Original post by TommyCricket)
Numerator is half the differential of the numerator so the integral is 1/2log(5x^2 + 7)
Yeah I've looked at the answers and thats right, I just don't know how to get there? I've been told to use this formula

$\int f(x)dx= \int f(x)\frac{dx}{du}du$
6. (Original post by Banny Dyrne)
Yeah the substitution is $u = 5x^{2}+7$
As tommy cricket says this is not the best way to integrate this but,
Find then
7. (Original post by Banny Dyrne)
Yeah the substitution is $u = 5x^{2}+7$
So your integral contains a 1/u factor and you need work out du/dx which hopefully will allow you to write the 5xdx bit simply in terms of du.
8. I'm really not getting this. Here's what I've worked out so far:

$\int \frac{5x}{5x^{2}+7}$

$\frac{dx}{du}=\frac{1}{10x}$

so using the formula I have...

$\int \frac{5x}{5x^2+7}(\frac{1}{10x})du$

and thats where I get lost.
9. Now simplify..
10. $\frac{5x}{50x^{3}+70x}du$
?
11. (Original post by Banny Dyrne)
I'm really not getting this. Here's what I've worked out so far:

$\int \frac{5x}{5x^{2}+7}$

$\frac{dx}{du}=\frac{1}{10x}$

so using the formula I have...

$\int \frac{5x}{5x^2+7}(\frac{1}{10x})du$

and thats where I get lost.
No no no no, first off you would (theoretically) be able to cancel out the x in the numerator and the denominator, but then you have a function of two dependant variables u and x, which is a pain in the arse to simplify.

You were right in thinking:

through some nifty rearranging you have:

Does that make things any simpler when substituting?
12. Not really! Haha. I think I'm going to go over everything from the beginning of this chapter in my textbook. I'm probably missing something so simple and I just can't get my head around it right now. Thanks for trying everyone and I'll let you know when I solve it!
13. (Original post by Banny Dyrne)
Not really! Haha. I think I'm going to go over everything from the beginning of this chapter in my textbook. I'm probably missing something so simple and I just can't get my head around it right now. Thanks for trying everyone and I'll let you know when I solve it!
You're substituting in u to make an integral for u.

Would you be able to integrate

?

?
14. Yeah,

$\frac{1}{2}\ln x+c$

and

$\frac{1}{2}\ln u+c$
15. (Original post by Banny Dyrne)
I'm really not getting this. Here's what I've worked out so far:

$\int \frac{5x}{5x^{2}+7}$

$\frac{dx}{du}=\frac{1}{10x}$

so using the formula I have...

$\int \frac{5x}{5x^2+7}(\frac{1}{10x})du$

and thats where I get lost.
Imagine 5x2 + 7 and 10x are swapped. Does that make it easier to see?
16. YES! Nice one. Also thanks to everyone who's spent the last three hours encouraging me.

Finally...

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