Integration by substitution.

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.

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

Reply 1

joostan

13

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?

Reply 2

TommyCricket

2

Numerator is half the differential of the denominator so the integral is 1/2log(5x^2 + 7)

(edited 11 years ago)

Reply 3

Banny Dyrne

OP

3

Yeah the substitution is $u = 5x^{2}+7$

Reply 4

Banny Dyrne

OP

3

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

$\int f(x)dx= \int f(x)\frac{dx}{du}du$

Reply 5

joostan

13

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

\frac{du}{dx}

then

\ 1 \div\frac{du}{dx} \ =\frac{dx}{du}

(edited 11 years ago)

Reply 6

davros

16

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.

Reply 7

Banny Dyrne

OP

3

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.

Now simplify..

OP

$\frac{5x}{50x^{3}+70x}du$
?

Reply 10

munn

16

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:

\frac{du}{dx}=10x

through some nifty rearranging you have:

{5x}\text{ dx}= \frac{1}{2}du

Does that make things any simpler when substituting?

Reply 11

Banny Dyrne

OP

3

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!

Reply 12

munn

16

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

\int{\frac{1}{2x}\text{dx}}

?

what about

\int{\frac{1}{2u}\text{du}}

?

Reply 13

Banny Dyrne

OP

3

Yeah,

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

and

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

Reply 14

Exon

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?