Integration

[Aladdin]

I need some help here guys , please .

Integrat the following :

$\displaystyle \int\frac{1}{x^2-6x+5}$

My work :

$\displaystyle \int \frac{A}{x-1}+\frac{B}{x-5}$ where A & B must be determined

Now would I say that (A)(x-5) + (B)(x-1) = 1 and calculate A & B

take x as a common factor and set the coeficient equal to zero .... etc

Thanks in advance
Aladdin

 

[Deleted member 4993]

I need some help here guys , please .

Integrat the following :

$\displaystyle \int\frac{1}{x^2-6x+5}$

My work :

$\displaystyle \int \frac{A}{x-1}+\frac{B}{x-5}$ where A & B must be determined

Now would I say that (A)(x-5) + (B)(x-1) = 1 and calculate A & B

take x as a common factor and set the coeficient equal to zero .... etc

Thanks in advance
Aladdin

There is a short-cut

$\displaystyle \frac{1}{x^2-6x+5} = \frac{A}{x-1} + \frac{B}{x-5}$.............................(1)

Multiply both sides by (x-1)

$\displaystyle \frac{1}{x-5}= A + \frac{B(x-1)}{x-5}$

Now find the limit of above expression as x ? 1, and we get:

$\displaystyle \frac{1}{1-5}= A$

$\displaystyle A = -\frac{1}{4}$

Now to find B, multiply (1) by (x-5) and find the limit as x?5.

This approach will not work in case of repeated roots.

 

[galactus]

Yep, you can do this.

$\displaystyle A(x-5)+B(x-1)=1$

$\displaystyle Ax-5A+Bx-B=1$

Equate ceofficeints:

$\displaystyle A+B=0$

$\displaystyle -5A-B=1$

Now,solve that little system and that's it.

 

[Aladdin]

Thank you ~

Thanks Mr khan , I loved your way but I'll use my way ~ Cody's ~

 

[BigGlenntheHeavy]

$\displaystyle \int\frac{dx}{x^{2}-6x+5} \ = \ \int\bigg[\frac{1}{4(x-5)}-\frac{1}{4(x-1)}\bigg]^{*}dx$

$\displaystyle = \ \frac{1}{4}\int\frac{dx}{x-5}-\frac{1}{4}\int\frac{dx}{x-1} \ = \ \frac{1}{4}ln|x-5|-\frac{1}{4}ln|x-1|+C$

$\displaystyle = \ \frac{1}{4}ln\bigg|\frac{x-5}{x-1}\bigg|+C$

$\displaystyle ^{*} \ TI-89, \ Expand, \ to \ wit: \ (F2-3) \ \bigg(\frac{1}{(x^{2}-6x+5)},x\bigg) \ = \ \frac{1}{4(x-5)}-\frac{1}{4(x-1)}$

 

[Aladdin]

Thanks Gleen ~ Nice work .

 

[BigGlenntheHeavy]

Aladdin, it's Glenn, not Gleen.

 

[BigGlenntheHeavy]

$\displaystyle Hint: \ A \ tip, \ b \ a \ constant, \ obviously \ b \ \ne \ 0.$

$\displaystyle \int\frac{du}{u^{2}-b} \ \implies \ Natural \ Logarithm \ Solution, \ \frac{1}{u^{2}-b} \ can \ be \ expanded.$

$\displaystyle \int\frac{du}{u^{2}+b} \ \implies \ Arctangent \ Solution, \ \frac{1}{u^{2}+b} \ is \ prime.$