Integration Question

**Karl Harder** · August 29th 2009, 03:09 AM

Is it fair to say that or am i wrong? not sure can someone verify please

$-\frac{5}{2}\int_{}^{t}\frac{dt}{3(t^2+a^2+at)} = -\frac{5}{2}ln|3(t^2+a^2+at)|$

**skeeter** · August 29th 2009, 03:21 AM

Originally Posted by Karl Harder

Is it fair to say that or am i wrong? not sure can someone verify please

$-\frac{5}{2}\int_{}^{t}\frac{dt}{3(t^2+a^2+at)} = -\frac{5}{2}ln|3(t^2+a^2+at)|$

take the derivative of your result ... does that get you back to the integrand?

**Karl Harder** · August 29th 2009, 04:00 AM

Ok so it probably won't work as differentiating the equation does not return me back to my integral. I'm not sure on how you integrate fractions very well. Any hits with different methods on how to integrate fractions and the rules that you use to determine the answer?

**Matt Westwood** · August 29th 2009, 04:18 AM

General techniques for solving an integral with a quadratic on the bottom:

a) If you can, factorise the quadratic, then use the technique of partial fractions to turn it into something of the form $\int \left({\frac A {x+a} + \frac B {x+b}}\right) dx$ where A, a, B, b are constants;

b) Turn the quadratic into the form $k \left({(x + a)^2 \pm b^2}\right)$ and then use a trigonometrical substitution. (This will also work in the first case - as for me, I prefer it as there's nothing more boringly tedious than partial fraction expansions and I avoid them whenever I can).

**skeeter** · August 29th 2009, 04:23 AM

Originally Posted by Karl Harder

Ok so it probably won't work as differentiating the equation does not return me back to my integral. I'm not sure on how you integrate fractions very well. Any hits with different methods on how to integrate fractions and the rules that you use to determine the answer?

different techniques are done dependent on the nature of the rational function.

for your integral ...

$-\frac{5}{2} \int \frac{dt}{3(t^2+at+a^2)}$

$-\frac{5}{6} \int \frac{dt}{t^2+at+a^2}$

$-\frac{5}{6} \int \frac{dt}{\left(t^2+at+ \frac{a^2}{4}\right) + \frac{3a^2}{4}}$

$-\frac{5}{6} \int \frac{dt}{\left(t + \frac{a}{2}\right)^2 + \frac{3a^2}{4}}$

$-\frac{5}{6} \int \frac{dt}{\left(t + \frac{a}{2}\right)^2 + \left(\frac{\sqrt{3}a}{2}\right)^2}$

now note that integrals of the form ...

$\int \frac{dx}{x^2+k^2} = \frac{1}{k}\arctan\left(\frac{x}{k}\right) + C$

**HallsofIvy** · August 29th 2009, 07:38 AM

Originally Posted by Karl Harder

Is it fair to say that or am i wrong? not sure can someone verify please

$-\frac{5}{2}\int_{}^{t}\frac{dt}{3(t^2+a^2+at)} = -\frac{5}{2}ln|3(t^2+a^2+at)|$

Essentially, you are ignoring the fact that $t^2+_ a^2+ at$ is a function of t and not t itself- and ignoring things is never a good idea!

If you start from [tex]-\frac{5}{2}ln|3(t^2+ a^2+ at)| and differentiate you will get, by the chain rule, $-\frac{5}{2}\frac{1}{3(t^2+ a^2+ at)}$ times the derivative of $3(t^2+ a^2+ at)$ . Because that is not just "t", that derivative is NOT 1.

That is why, in "integration by substitution" requires that, if you substitute "u= f(t)" you must already have f'(t) in the integral so you can also substitute "du= f'(t) dt".

Matt Westwood and skeeter have already made suggestions on integrating. I am pointing out why what you are trying to do doesn't work. Oddly enough, if this were the "more complicated" integral
$-\frac{5}{2}\int \frac{(2t+ a)dt}{3(t^2+a^2+at)}$
that would work. You could make the substitution $u= t^2+ a^2+ at$ and the derivative, $du= (2t+ a)dt$ is already in the integral.

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