# Math Help - Partial Fraction /Trig Integral

1. ## Partial Fraction /Trig Integral

$\displaystyle \int \frac{x dx}{x^2 + 4x + 13}$

Completing the square,
$\displaystyle \int \frac{x dx}{(x+2)^2 + 9}$

$let u = x + 2; x = u-2; du = dx$

$\int\frac{(u-2)du}{u^2 + 3^2}$

$\int\frac{u du}{u^2 + 3^2} - 2\int\frac{du}{u^2 + 3^2}$

$let v = u^2 + 9; dv = 2u du$

$\frac{1}{2}\int \frac{dv}{v} - \frac{2}{3}\arctan(\frac{u}{3})$

This is actually a definite integral, but I'd like to make sure that I have the antiderivative done correctly, since that is most likely where I have screwed up. Does anybody see any mistakes here?

Thanks!

2. ## Re: Partial Fraction /Trig Integral

Originally Posted by joatmon
$\displaystyle \int \frac{x dx}{x^2 + 4x + 13}$

Completing the square,
$\displaystyle \int \frac{x dx}{(x+2)^2 + 9}$

$let u = x + 2; x = u-2; du = dx$

$\int\frac{(u-2)du}{u^2 + 3^2}$

$\int\frac{u du}{u^2 + 3^2} - 2\int\frac{du}{u^2 + 3^2}$

$let v = u^2 + 9; dv = 2u du$

$\frac{1}{2}\int \frac{dv}{v} - \frac{2}{3}\arctan(\frac{u}{3})$

This is actually a definite integral, but I'd like to make sure that I have the antiderivative done correctly, since that is most likely where I have screwed up. Does anybody see any mistakes here?

Thanks!
\displaystyle \begin{align*} \int{\frac{x}{x^2 + 4x + 13}\,dx} &= \frac{1}{2}\int{\frac{2x}{x^2 + 4x + 13}} \\ &= \frac{1}{2}\int{\frac{2x + 4}{x^2 + 4x + 13}\,dx} - \frac{1}{2}\int{\frac{4}{x^2 + 4x + 13}\,dx} \\ &= \frac{1}{2}\int{\frac{2x+4}{x^2 + 4x + 13}\,dx} - 2\int{\frac{1}{(x + 2)^2 + 9}\,dx}\end{align*}

For the first make the substitution $\displaystyle u = x^2 + 4x + 13 \implies du = (2x + 4)\,dx$ and for the second make the substitution $\displaystyle x + 2 = 3\tan{\theta} \implies dx = 3\sec^2{\theta}\,d\theta$.

3. ## Re: Partial Fraction /Trig Integral

OK. I see what you are doing, but I'm still stuck. Focusing on the 2nd substitution, I get this:

$6 \int \frac{\sec^2\theta}{(3\tan^2\theta) + 3^2}d\theta$

But then I don't know what to do. This looks like a tangent integral, but I am used to either a constant in the numerator or the denominator being inside of a square root. I don't know how to handle this.

Can you help me further please. I appreciate it. Thanks.

4. ## Re: Partial Fraction /Trig Integral

Originally Posted by joatmon
OK. I see what you are doing, but I'm still stuck. Focusing on the 2nd substitution, I get this:

$6 \int \frac{\sec^2\theta}{(3\tan^2\theta) + 3^2}d\theta$

But then I don't know what to do. This looks like a tangent integral, but I am used to either a constant in the numerator or the denominator being inside of a square root. I don't know how to handle this.

Can you help me further please. I appreciate it. Thanks.
The denominator should actually be $\displaystyle (3\tan{\theta})^2 + 3^2 = 9\tan^2{\theta} + 9 = 9\left(\tan^2{\theta} + 1\right) = 9\sec^2{\theta}$.

This will simplify the integrand considerably...

5. ## Re: Partial Fraction /Trig Integral

Originally Posted by joatmon
OK. I see what you are doing, but I'm still stuck. Focusing on the 2nd substitution, I get this:

$6 \int \frac{\sec^2\theta}{(3\tan^2\theta) + 3^2}d\theta$

But then I don't know what to do. This looks like a tangent integral, but I am used to either a constant in the numerator or the denominator being inside of a square root. I don't know how to handle this.

Can you help me further please. I appreciate it. Thanks.
It should be $6\int\frac{\sec^2\theta}{(3\tan\theta)^2+3^2}\, d\theta$.

Now observe that $(3\tan\theta)^2+3^2=9\tan^2\theta+9=9(\tan^2\theta +1)=9\sec^2\theta$

Something nice happens now...what is it?