1. ## Integration problems

I can't figure out how to integrate the following

$2/4+x^2$

$-1/(x^2+4x+14)$

I've tried them both, and on the first one, I can't get past the initial steps of u=4+x^2, du=2x.

On the second, I assume you have to complete the square at some point, but I don't know what to do beyond that.

2. Actually, in both cases, you make use of the fact that the antiderivative of $\frac{1}{1+x^2}$ is $\arctan{x}+C$. Thus $\int\frac{dx}{a^2+x^2}$ is, via the u-substitution $u=\frac{x}{a}$, found to be:
$\int\frac{dx}{a^2+x^2}=\int\frac{a\,du}{a^2+a^2u^2 }$
$\,\,=\frac{1}{a}\int\frac{du}{1+u^2}$
$\,\,=\frac{1}{a}\arctan\frac{x}{a}+C$

Both components of your integrand should be able to be transformed into multiples of $\frac{1}{a^2+u^2}$ forms the above using appropriate u-substitutions.
-Kevin C.

3. Originally Posted by quarks
[snip]
$-1/(x^2+4x+14)$

[snip]

On the second, I assume you have to complete the square at some point, but I don't know what to do beyond that.
$-\frac{1}{x^2+4x+14} = - \frac{1}{(x + 2)^2 + 10}$.

Now make the substitution u = x + 2. Now take another look at the previous twisted reply.