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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: squance on July 30, 2008, 07:39:50 pm

Title: Integration...NEED HELP!!! PLZ!!
Post by: squance on July 30, 2008, 07:39:50 pm: These are some integration questions I recently faced in my first tutorial class at uni.
It was supposed to be revision and we were expected to know this...
except...I forgotten how to do these.

Can someone help please?
Solutions are at the bottom.
Thanks

(h) $\int\frac{1}{x^2+2x+6}dx$

(i) $\int\frac{x}{\sqrt{1+x}}dx$

(j) $\int\frac{x^4-3x^2}{x^5-5x^3-24}dx$

(k) $\int\frac{4x}{(2x-1)^2}dx$

[IMG]http://img167.imageshack.us/img167/794/img0001oq2.th.jpg[/img]
Title: Re: Integration...NEED HELP!!! PLZ!!
Post by: ell on July 30, 2008, 08:00:57 pm: Couple of hints:

(h) Complete the square for the denominator, $x^2 + 2x + 6 = x^2 + 2x + 1 + 5 = (x+1)^2 + 5$ . Then apply $\int \frac{a}{a^2 + x^2}\ dx = \tan^{-1} \frac{x}{a} + C$

(i) Use the substitution $u = 1 + x$

Then $x = u - 1$ and $\frac{du}{dx} = 1$

(j) Let $u = x^5 - 5x^3 - 24$

Then $\frac{du}{dx} = 5x^4 - 15x^2 = 5(x^4 - 3x^2)$

The new integral is now $\frac{1}{5} \int \frac{1}{u}\ du$

(k) Another linear substitution - hopefully you'll be able to get this one, it is very similar to (i). :)
Title: Re: Integration...NEED HELP!!! PLZ!!
Post by: squance on July 30, 2008, 08:14:31 pm: Thanks.
I've have a go of them now.
Title: Re: Integration...NEED HELP!!! PLZ!!
Post by: squance on July 31, 2008, 11:51:56 am: So for the first one...

(h) $\int\frac{1}{x^2+2x+6}dx$

completing square of denominator is $(x+1)^2+5$

so.... $\int\frac{1}{(x+1)^2+5}dx$

Let u = x + 1 du/dx = 1
x=u-1

so $\int\frac{1}{u^2+5}du$

= $\frac{1}{\sqrt5}arctan\frac{u}{\sqrt{5}}+c$

= $\frac{1}{\sqrt5}arctan\frac{x+1}{\sqrt{5}}+c$

(i) $\int\frac{x}{\sqrt{1+x}}dx$
Let u = 1 + x du/dx = 1
x = u-1

$\int\frac{u-1}{\sqrt{1+u-1}}du$
$\int\frac{u-1}{\sqrt{u}}du$
$\int u^\frac{1}{2} - u^\frac{-1}{2} du$
$\frac{2}{3} (u)^\frac{3}{2}-2(u)^\frac{1}{2}$
$\frac{2}{3} (1+x)^\frac{3}{2}-2(1+x)^\frac{1}{2}+c$

Is this correct?
(I can't be stuffed putting up the other two but I'll be happy if someone can tell me if these two are right...)
Title: Re: Integration...NEED HELP!!! PLZ!!
Post by: dcc on July 31, 2008, 04:21:48 pm: Both are correct :)