Help with understanding this u-substitution that makes no sense to me

Question

$ \int{\frac{x^2+4x+4}{(x+1)^2(x^2+3x+3)}}dx$

The question: Integrate the above, using partial fraction decomposition.

Getting the partial fraction is fine:

$=-{\displaystyle\int}\dfrac{x+2}{x^2+3x+3}\,\mathrm{d}x+{\displaystyle\int}\dfrac{1}{x+1}\,\mathrm{d}x+{\displaystyle\int}\dfrac{1}{\left(x+1\right)^2}\,\mathrm{d}x$

The problem: After doing the partial fraction decomposition, there is an integral that needs to be evaluated using u-substitution:

${\displaystyle\int}\dfrac{1}{x^2+3x+3}\,\mathrm{d}x$

Mathematica and other solvers all give this u-substitution:

Complete the square:

$={\displaystyle\int}\dfrac{1}{\left(x+\frac{3}{2}\right)^2+\frac{3}{4}}\,\mathrm{d}x$

Substitute $u=\dfrac{2x+3}{\sqrt{3}} \longrightarrow \dfrac{\mathrm{d}u}{\mathrm{d}x}=\dfrac{2}{\sqrt{3}}$ ( <--WTF?!)

$=\class{steps-node}{\cssId{steps-node-6}{\dfrac{2}{\sqrt{3}}}}{\displaystyle\int}\dfrac{1}{u^2+1}\,\mathrm{d}u$

I have no idea why or how this substitution works and more importantly, the approach one would take it to come up with it. This is driving me nuts. My lecturers are useless.

S.O.S

Answer 1

Note that $$\left(x+\frac 32\right)^2+\frac 34=\frac 34\left(\frac 43\left(x+\frac 32\right)^2+1\right)=\frac 34\left(\left(\frac {2x}{\sqrt 3}+\frac 3{\sqrt 3}\right)^2+1\right)$$You are looking for a substitution that gets you $u^2+1$ hence taking out the factor $\frac 34$ and then taking the $\frac 43$ into the bracket. The suggested substitution gets you to $$\frac 34(u^2+1)$$ for the original integrand, and you also need to take the derivative into account, as always with substitutions.

Answer 2

I would have done the substitution in two steps. The goal is to make the denominator look like $x^2+1$. The steps for this are always the same: 1. Complete the square to get rid of the middle term. 2. Make the constant term equal to 1. 3. Make the coefficient on the variable equal to one.

Step one, let $v=x+3/2$ so that $dv=dx$. Now the integral looks like

$$\int \frac{1}{v^2+\frac{3}{4}} \; dv.$$

Step two, factor $3/4$ out of the denominator. Heck, take it clear outside the integral.

$$\frac{4}{3} \int \frac{1}{\frac{4}{3}v^2+1} \; dv.$$

Step three, put the coefficient on $v$ inside the square:

$$\frac{4}{3} \int \frac{1}{\left(\frac{2}{\sqrt{3}}v\right)^2+1} \; dv.$$

and then substitute $u=2v/\sqrt{3}$, so that $du = (2/\sqrt{3})\; dv$:

$$\frac{4}{3} \int \frac{1}{u^2+1} \frac{\sqrt{3}}{2} \; du.$$

Push the constant outside to get:

$$\frac{2}{\sqrt{3}} \int \frac{1}{u^2+1} \; du.$$

Now trace back the substitutions:

$$u = \frac{2}{\sqrt{3}} v = \frac{2}{\sqrt{3}} \left(x+\frac{3}{2}\right)=\frac{2x+3}{\sqrt{3}}.$$

Mathematicians have a great tendency to hide their scratch paper and act like they did everything in one line.