# Integration by substitution.

• Feb 11th 2013, 03:07 PM
Furyan
Integration by substitution.
Hello I'm sorry to ask again, but would somone be willing to look over this and tell me where I've gone wrong. Thank you.

I know the method I'm using is probably very shoddy, but it's the best I could do.

The question is integrate the following using the given substitution.

$\displaystyle \int\dfrac{\sqrt{x^2 + 4}}{x} dx$

$\displaystyle u^2 = x^2 + 4$

I got $\displaystyle \dfrac{dx}{du} = \dfrac{u}{x}$

I substituted these in first and got:

$\displaystyle \int(\dfrac{u}{x})(\dfrac{u}{x}) du$

$\displaystyle \int\dfrac{u^2}{x^2}$

Then using $\displaystyle u^2 = x^2 + 4$, I substituted again and got.

$\displaystyle \int\dfrac{u^2}{u^2 - 4}$

I didn't know how to integrate that so I used partial fractions and got:

$\displaystyle \int1 + \dfrac{1}{u - 2} - \dfrac{1}{u + 2}$

$\displaystyle u + \ln(u - 2) - \ln(u + 2) + c$

Using $\displaystyle u = \sqrt{x^2+4}$, I got:

$\displaystyle \sqrt{x^2 + 4} + \ln\dfrac{\sqrt{x^2+4} -2}{\sqrt{x^2 + 4} + 2} + c$

$\displaystyle \sqrt{x^2 + 4} - \ln\dfrac{\sqrt{x^2+4} + 2}{\sqrt{x^2 - 4} - 2} + c$, which according to my graphing calculator is very different. I understand that this might be too much too ask, especially as I have only shown an outline of my working, but if anyone one can see an obvious place we're I've gone wrong I'd really appreciate it. My graphing calulator tells me the partial fractions are correct, so I really don't see what is wrong with what I have done.

Thank you.
• Feb 11th 2013, 03:12 PM
Plato
Re: Integration by substitution.
Quote:

Originally Posted by Furyan
Hello I'm sorry to ask again, but would somone be willing to look over this and tell me where I've gone wrong. Thank you.

I know the method I'm using is probably very shoddy, but it's the best I could do.

The question is integrate the following using the given substitution.

$\displaystyle \int\dfrac{\sqrt{x^2 + 4}}{x} dx$

$\displaystyle u^2 = x^2 + 4$

I got $\displaystyle \dfrac{dx}{du} = \dfrac{u}{x}$

I substituted these in first and got:

$\displaystyle \int(\dfrac{u}{x})(\dfrac{u}{x}) du$

$\displaystyle \int\dfrac{u^2}{x^2}$

Then using $\displaystyle u^2 = x^2 + 4$, I substituted again and got.

$\displaystyle \int\dfrac{u^2}{u^2 - 4}$

I didn't know how to integrate that so I used partial fractions and got:

$\displaystyle \int1 + \dfrac{1}{u - 2} - \dfrac{1}{u + 2}$

$\displaystyle u + \ln(u - 2) - \ln(u + 2) + c$

Using $\displaystyle u = \sqrt{x^2+4}$, I got:

$\displaystyle \sqrt{x^2 + 4} + \ln\dfrac{\sqrt{x^2+4} -2}{\sqrt{x^2 + 4} + 2} + c$

$\displaystyle \sqrt{x^2 + 4} - \ln\dfrac{\sqrt{x^2+4} + 2}{\sqrt{x^2 - 4} - 2} + c$, which according to my graphing calculator is very different. I understand that this might be too much too ask, especially as I have only shown an outline of my working, but if anyone one can see an obvious place we're I've gone wrong I'd really appreciate it. My graphing calulator tells me the partial fractions are correct, so I really don't see what is wrong with what I have done.

Thank you.

Look at this.
• Feb 11th 2013, 03:39 PM
Furyan
Re: Integration by substitution.
Hello Plato,

Well apart from the fact that they all start with $\displaystyle \sqrt{x^2 + 4}$, they are all quite different. Well it's some consolation that Wolfram's answer is different to the given answer, but then I still don't see where I've gone wrong (Crying). Just out of interest why is Wolfram using $\displaystyle \log_{10}$.

Apparently 'small animal care' is under-subscribed at my college, I think I might migrate.(Sadsmile)
• Feb 11th 2013, 03:52 PM
Plato
Re: Integration by substitution.
Quote:

Originally Posted by Furyan
Well apart from the fact that they all start with $\displaystyle \sqrt{x^2 + 4}$, they are all quite different. Well it's some consolation that Wolfram's answer is different to the given answer, but then I still don't see where I've gone wrong (Crying). Just out of interest why is Wolfram using $\displaystyle \log_{10}$.

They are not using $\displaystyle \log_{10}$.

Like most modern practice in mathematics we use $\displaystyle \log(x)$ in place of the out-of-date $\displaystyle \ln(x)$.

See a good calculus textbook like Gillman & McDowell.
• Feb 11th 2013, 04:12 PM
Furyan
Re: Integration by substitution.
Right. Thank you for the book recommendation.

I have just checked the three solutions again. The one I got is, in fact, equivalent to the one given by Wolfram (Party)

Thank you for the link to that solution.
• Feb 12th 2013, 02:16 AM
tom@ballooncalculus
Re: Integration by substitution.
More to the point, look here.

I.e. your substitution is fine (though completely different from Wolfram's), just see 'alternate forms' no. 1. (Multiplying top and bottom by the top.)

Probably the sign typo here, though...

Quote:

Originally Posted by Furyan
$\displaystyle \sqrt{x^2 + 4} - \ln\dfrac{\sqrt{x^2+4} + 2}{\sqrt{x^2 - 4} - 2} + c$