Integral (shouldn't be too hard)

**U-God** · October 29th 2008, 07:21 PM

Hey guys, would be nice if I could have help with a quick integral.

$\int \frac{1}{x^2 - 5x + 4} dx$

I'm pretty sure that I need to make a substitution here, but I can't see how.

Cheers,

**tsal15** · October 29th 2008, 07:57 PM

Originally Posted by U-God

Hey guys, would be nice if I could have help with a quick integral.

$\int \frac{1}{x^2 - 5x + 4} dx$

I'm pretty sure that I need to make a substitution here, but I can't see how.

Cheers,

Hey U-God,

I got this,

$\int \frac{1}{x^2 - 5x + 4} dx$

$\frac{1}{3} \int \frac{3}{x^2 - 5x + 4} dx$

$\frac{1}{3} \int \frac{3}{(x-1)(x-4)} dx$

$\frac{1}{3} \int \frac{-1}{(x-1)} + \frac{1}{(x-4)} dx$

$\frac{1}{3} (\ln{(x-4)} - \ln{(x-4)})$

I didn't see the need for any substitution... Its perfectly fine without any substitution

Hopefully I've been helpful.

tsal15

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