Integration

A

Aladdin

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Mar 27, 2009
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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
 
Aladdin said:
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
Click to expand...
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.
 
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.
 
Thank you ~

Thanks Mr khan , I loved your way but I'll use my way :) ~ Cody's ~
 
\(\displaystyle \int\frac{dx}{x^{2}-6x+5} \ = \ \int\bigg[\frac{1}{4(x-5)}-\frac{1}{4(x-1)}\bigg]^{*}dx\)

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

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

\(\displaystyle ^{*} \ TI-89, \ Expand, \ to \ wit: \ (F2-3) \ \bigg(\frac{1}{(x^{2}-6x+5)},x\bigg) \ = \ \frac{1}{4(x-5)}-\frac{1}{4(x-1)}\)
 
\(\displaystyle Hint: \ A \ tip, \ b \ a \ constant, \ obviously \ b \ \ne \ 0.\)

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

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