# Conceptual Integrals

• Feb 25th 2010, 12:31 AM
Selim
Conceptual Integrals
$\int h(x)f^1(x) * dx = \int$

and

$\int \frac{2}{3x^2 + 4x + 5} * dx = \int
$
• Feb 25th 2010, 12:48 AM
Pulock2009
Quote:

Originally Posted by Selim
$\int h(x)f^1(x) * dx = \int$

and

$\int \frac{2}{3x^2 + 4x + 5} * dx = \int
$

your second question is very easy. express the denominator in the form x^2+-a^2 where a is a constant. then integrate.
• Feb 25th 2010, 02:03 AM
Prove It
Quote:

Originally Posted by Selim
$\int h(x)f^1(x) * dx = \int$

and

$\int \frac{2}{3x^2 + 4x + 5} * dx = \int
$

The bottom does not factorise, so you can't use partial fractions.

That means you'll have to complete the square and use a trigonometric substitution.

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

$= \frac{2}{3}\int{\frac{1}{x^2 + \frac{4}{3}x + \left(\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2 + \frac{5}{3}}\,dx}$

$= \frac{2}{3}\int{\frac{1}{\left(x + \frac{2}{3}\right)^2 + \frac{11}{9}}\,dx}$

Now make the substitution $x + \frac{2}{3} = \sqrt{\frac{11}{9}}\tan{\theta} = \frac{\sqrt{11}}{3}\tan{\theta}$.

Note that $dx = \frac{\sqrt{11}}{3}\sec^2{\theta}\,d\theta$.

Also note that $\theta = \arctan{\left[\frac{3\sqrt{11}}{11}\left(x + \frac{2}{3}\right)\right]}$.

Then the integral becomes

$= \frac{2}{3}\int{\frac{1}{\left(\frac{\sqrt{11}}{3} \tan{\theta}\right)^2 + \frac{11}{9}}\,\frac{\sqrt{11}}{3}\sec^2{\theta}\, d\theta}$

$= \frac{2}{3}\int{\frac{\frac{\sqrt{11}}{3}\sec^2{\t heta}}{\frac{11}{9}\tan^2{\theta} + \frac{11}{9}}\,d\theta}$

$= \frac{2}{3}\int{\frac{\frac{\sqrt{11}}{3}\sec^2{\t heta}}{\frac{11}{9}(\tan^2{\theta} + 1)}\,d\theta}$

$= \frac{2\sqrt{11}}{11}\int{\frac{\sec^2{\theta}}{\s ec^2{\theta}}\,d\theta}$

$= \frac{2\sqrt{11}}{11}\int{1\,d\theta}$

$= \frac{2\sqrt{11}}{11}\theta + C$

$= \frac{2\sqrt{11}}{11}\arctan{\left[\frac{3\sqrt{11}}{11}\left(x + \frac{2}{3}\right)\right]} + C$.
• Feb 25th 2010, 12:12 PM
Selim
Thanks for that messy one, sir, and the other one is pretty easy,

$\int h(x) * f^1(x) * dx = h(x) * f(x) - \int f(x) * h^1(x)$

I was simply supposed to integrate it by parts.