How to integrate $\frac{1}{(x^2-4x+6)^{3/2}}$ with respect to x?

Question

I need to find $\int\frac{1}{(x^2-4x+6)^{3/2}}dx$ but am not sure where to start. I tried breaking up the denominator into $(x^2-4x+6)$ and $(x^2-4x+6)^\frac1{2}$ but can't figure out how I would integrate that by parts. I also thought about using partial fractions, but I don't know how to do that with a fractional exponent.

Answer 1

The first thing you think of when you see $x^2-4x+6$ is completing the square: $$ (x^2-4x + 4) + 2 = (x-2)^2+2 = \underbrace{{}\quad u^2 + 2\quad {}}_\text{No 1st-degree term}. $$ Then you have $$ \int \frac{du}{(u^2+2)^{3/2}}\tag 1 $$ (where $dx$ has become $du$ since $\dfrac d {dx} (x-2)=1$).

Letting $u = \sqrt 2 \tan\theta$ we get $du = \sqrt 2 \sec^2\theta\,d\theta$ and $u^2+2 = 2\sec^2\theta$.

The purpose of completing the square is always to reduce a problem involving a quadratic polynomial with a first-degree term to a problem involving a quadratic polynomial with no first-degree term.

The integral $(1)$ becomes $$ \int \frac{\sqrt 2 \sec^2\theta\,d\theta}{\sqrt 2 ^3 \sec^3\theta} = \frac 1 2 \int \frac{d\theta}{\sec\theta}. $$ That one is found in all the tables.

At some point you'll need to convert $\sec\theta$ back to $\sqrt{u^2+2}$ and $\tan\theta$ to $u/\sqrt 2$, and then convert back to functions of $x$. (Unless you're doing a definite integral, in which case, just change to the appropriate bounds on $\theta$ and plug those in for $\theta$ at the end.)