ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Glockmeister on May 04, 2008, 10:15:28 pm

Title: Problem
Post by: Glockmeister on May 04, 2008, 10:15:28 pm: $\int \frac{x+2}{x^{2}+1} dx$

How would you do it?
Title: Re: Problem
Post by: Mao on May 04, 2008, 10:24:50 pm: $\int \frac{x+2}{x^2+1} dx = \int \frac{x}{x^2+1} dx + 2\int \frac{1}{x^2+1}dx$

looking at the first integral, using the method of substitution:

let $u=x^2+1$ , $\implies dx = \frac{du}{2x}$

$\int \frac{x}{x^2+1} dx=\int \frac{x}{u} \cdot \frac{du}{2x}$

$\int \frac{x}{x^2+1} dx=\frac{1}{2}\int \frac{1}{u}du$

$\int \frac{x}{x^2+1} dx=\frac{1}{2}log_e\left(x^2+1\right) +c$

looking at the second integral, if you have memorised your inverse trig derivatives, you'll notice that:

$\frac{d}{dx}\tan^{-1}x = \frac{1}{x^2+1}$

$\implies \int \frac{1}{x^2+1} dx = \tan^{-1}x +c$

hence

$\int \frac{x+2}{x^2+1} dx = \frac{1}{2}log_e\left(x^2+1\right) + 2\tan^{-1}x +c$
Title: Re: Problem
Post by: Glockmeister on May 04, 2008, 11:32:14 pm: That's pretty smart actually, thanks for that Mao
Title: Re: Problem
Post by: AppleXY on May 09, 2008, 07:37:51 am: yeah best to use linear subsitution for these probs :)