factorise trouble

Hedgecutter

I need to factorise 6x^2+5, I think I m allowed put in ox to factorise but cant go any further.

6x^2+0x+5

TheBody

Hedgecutter wrote: »

I need to factorise 6x^2+5, I think I m allowed put in ox to factorise but cant go any further.

6x^2+0x+5

Hi there. Unless you are talking about using complex numbers, it's can't be factorised over the real numbers.

Hedgecutter

TheBody wrote: »

Hi there. Unless you are talking about using complex numbers, it's can't be factorised over the real numbers.

ok its an integration problem I have: integrate (12/6x^2+5)dx

I was using partial fractions method

How do I use complex numbers?

TheBody

Hedgecutter wrote: »

ok its an integration problem I have: integrate (12/6x^2+5)dx

I was using partial fractions method

How do I use complex numbers?

Try tan substitution. I'll give you more help if you need it. Give it a go yourself first.

Hedgecutter

TheBody wrote: »

Try tan substitution. I'll give you more help if you need it. Give it a go yourself first.

Thanks I give that ago and reply later, have to take the kids to the parade.

thanks again.

TheBody

Hedgecutter wrote: »

Thanks I give that ago and reply later, have to take the kids to the parade.

thanks again.

Your welcome. Let us know how you get on with the problem.

Enjoy the parade!

Hedgecutter

TheBody wrote: »

Your welcome. Let us know how you get on with the problem.

Enjoy the parade!

http://mathsfirst.massey.ac.nz/Calculus/integration/IntSubstitution/IntSub3.html

Before I go is the first example what your talking about?

TheBody

Hedgecutter wrote: »

http://mathsfirst.massey.ac.nz/Calculus/integration/IntSubstitution/IntSub3.html

Before I go is the first example what your talking about?

That's it. Depending on the level you require, some lecturers are happy with people just using the identity,

[latex] \int{\frac{1}{x^2+a^2}dx=\frac{1}{a}(\tan^{-1}\frac{x}{a})+c[/latex]

With this in mind, you could rewrite your problem as:

[latex]\int{\frac{12}{6x^2+5}dx=12\int{\frac{1}{6(x^2+\frac{5}{6})}}dx=2\int{\frac{1}{x^2+\frac{5}{6}}}dx=2\int{\frac{1}{x^2+(\frac{\sqrt{5}}{\sqrt{6}})^2}}dx[/latex]

Using the identity above, can you finish it from there?

Hedgecutter

TheBody wrote: »

That's it. Depending on the level you require, some lecturers are happy with people just using the identity,

[latex] \int{\frac{1}{x^2+a^2}dx=\frac{1}{a}\tan^{-1}\frac{x}{a}[/latex]

With this in mind, you could rewrite your poblem as:

[latex]\int{\frac{12}{6x^2+5}dx=12\int{\frac{1}{6(x^2+\frac{5}{6})}}dx=2\int{\frac{1}{x^2+\frac{5}{6}}}dx=2\int{\frac{1}{x^2+(\frac{\sqrt{5}}{\sqrt{6}})^2}}dx[/latex]

Using the identity above, can you finish it from there?

Cant write using latex, hopefully you ll be able to make out what im saying

2 * 1/(square root5/6)^2 Tan-1 x/square root (5/6)?

TheBody

Hedgecutter wrote: »

Cant write using latex, hopefully you ll be able to make out what im saying

2 * 1/(square root5/6)^2 Tan-1 x/square root (5/6)?

Almost! That square shouldn't be there. Also, it's usual to simplify your answer.

What you have is:

[latex]2\int\frac{1}{x^2+(\frac{\sqrt{5}}{\sqrt{6}})^2}dx=2\frac{1}{\frac{\sqrt{5}}{\sqrt{6}}}\tan^{-1}\frac{x}{\frac{\sqrt{5}}{\sqrt{6}}}+c=2\frac{\sqrt{6}}{\sqrt{5}}}\tan^{-1}\frac{\sqrt{6}x}{\sqrt{5}}+c[/latex]

Hedgecutter

TheBody wrote: »

Almost! That square shouldn't be there. Also, it's usual to simplify your answer.

What you have is:

[latex]2\int\frac{1}{x^2+(\frac{\sqrt{5}}{\sqrt{6}})^2}dx=2\frac{1}{\frac{\sqrt{5}}{\sqrt{6}}}\tan^{-1}\frac{x}{\frac{\sqrt{5}}{\sqrt{6}}}+c=2\frac{\sqrt{6}}{\sqrt{5}}}\tan^{-1}\frac{\sqrt{6}x}{\sqrt{5}}+c[/latex]

that's my answer?

TheBody

Hedgecutter wrote: »

that's my answer?

Yes, that's it.

Hedgecutter

TheBody wrote: »

Yes, that's it.

Thanks appreciate your help, part (b) next to find the area between the line and a curve. might leave that till later.

TheBody

Hedgecutter wrote: »

Thanks appreciate your help, part (b) next to find the area between the line and a curve. might leave that till later.

Glad I could help.