How to do this hard integral. Partial fraction (1 Viewer)

[turntaker]

Plez help.

 

[rumbleroar]

Which one?

 

[Kurosaki]

Use partial fractions =).

 

[Kurosaki]

But yes, which one?

 

[turntaker]

Which one?

But yes, which one?

Sorry guys. Its 1, c

I'm such idiot

 

[Drongoski]

By inspection:

 

[Kurosaki]

Sorry guys. Its 1, c

I'm such idiot

Don't be so hard on yourself, most of us were pretty bad starting out .
Generally when you decompose partial fractions, you let the expression in the numerator be of a degree one less than that of the denominator; so as the denominators are quadratics the numerators are linear.

, and find the unknown coefficients.

Once you've gotten a bit more practice down you'll be able to see stuff like so - consider , and then just multiply a suitable constant in , and can bypass the above entirely (unless they ask you specifically to find the coefficients for the fraction's numerators in decomposed form).

 

[turntaker]

Don't be so hard on yourself, most of us were pretty bad starting out .
Generally when you decompose partial fractions, you let the expression in the numerator be of a degree one less than that of the denominator; so as the denominators are quadratics the numerators are linear.

, and find the unknown coefficients.

Once you've gotten a bit more practice down you'll be able to see stuff like so - consider , and then just multiply a suitable constant in , and can bypass the above entirely (unless they ask you specifically to find the coefficients for the fraction's numerators in decomposed form).

Thanks man. I just can't see it in my head. It seems magical to me that you can do that.

does it mean I should drop ext 2.

 

[Kurosaki]

Thanks man. I just can't see it in my head. It seems magical to me that you can do that.

does it mean I should drop ext 2.

No it means you need more practice . I think that you should see how you fare after you've done your first assessment and see what happens from there. If you do well, you should keep it; if you do OK but think you could have done better, likewise, and if you did really bad don't hesitate to drop. It's all up to you though.

 

[iStudent]

Don't be so hard on yourself, most of us were pretty bad starting out .
Generally when you decompose partial fractions, you let the expression in the numerator be of a degree one less than that of the denominator; so as the denominators are quadratics the numerators are linear.

, and find the unknown coefficients.

Once you've gotten a bit more practice down you'll be able to see stuff like so - consider , and then just multiply a suitable constant in , and can bypass the above entirely (unless they ask you specifically to find the coefficients for the fraction's numerators in decomposed form).

There's a slightly easier way. Since there are no 'x' terms and only constants and x^2 terms you don't have to have the numerator as Ax+B and Cx+D.

Simply having them as A and B is fine. As for the reason... well it's like letting u=x^2 so that you have linear products for your denominator and solving it as usual.

 

[Kurosaki]

There's a slightly easier way. Since there are no 'x' terms and only constants and x^2 terms you don't have to have the numerator as Ax+B and Cx+D.

Simply having them as A and B is fine. As for the reason... well it's like letting u=x^2 so that you have linear products for your denominator and solving it as usual.

Yes, I know that, but I felt showing the generalised case might be better for turntaker given they seem to be having trouble either understanding the process - how it works - or are simply not practised enough .

Nice explanation btw, hopefully turntaker will find it useful if they read over it.