# Partial fraction decomposition and trig substitution

I am reviewing material for the CAA module 0 exam and part of it is integration. I had trouble doing a partial fractions problem and the work that they having showing the answer is convoluted and unclear.

I am asked to integrate:

$\int \frac{x+7}{x^2(x+2)}dx$

obviously when you set it up you get:

$\frac{A}{x}+\frac{B}{x^2}+\frac{c}{x+2}=\frac{x+7}{x^2(x+2)}\tag{$*$}$

I had no problems here I had problems in the next step.

multiplying through I got:

$A(x^2)(x+2)+B(x)(x+2)+C(x^2)(x)=x+7$

Somehow they got $Ax(x+2)+B(x+2)+C(x^2)=x+7$

That is not even possible unless you factor out an x on the left hand side and divide both sides by x which would give you $\frac{x+7}{x}$ on the right hand side. So either their work is incorrect or I am missing something.

I also tried solving a trig substitution problem: (ignoring the constants they get pulled out anyway)

$\int\frac{1}{x^4\sqrt{x^2-7}}dx$

obviously $x=\sqrt{7}\sec(\theta)$

and eventually you get to: $\int \cos^3(\theta) dx$

peel off a cos: $\int \cos^2(\theta)\cos(\theta)\,d(\theta)$

use an identity:

$\int 1-\sin^2(\theta) \cos(\theta)$

let $u=\sin(\theta),\enspace du=\cos(\theta)\,d(\theta) \to d(\theta)=\frac{1}{\cos(\theta)}du$

which gives an integral of:

$\int 1-u^2 du$ because cos gets canceled out with substituting $d(\theta)$

which comes out to be $u-\frac{1}{3}u^3+c$

now in the reference triangle the Adjacent side $= \sqrt{7}$ Hypotenuse $=x$

Now is this right? This has to be right integration wise because even the answer showed an integral of cos cubed but then it went off and used some integration technique other than what I used which is the only sensible technique to peel off the cos. Their answer gave something retarded like $\frac{(2x^2+7)\sqrt{x^2-7}}{x^2 \times x}$ from the integral of $\cos^3(\theta)$ which came out to be $2+\cos^2(\theta)\sin(\theta)$. Am I going crazy? My method gives a valid solution, correct? and they used an unnecessary method right? I mean I was taught this method way back in calc II and I am going to be very frustrated if for some magical reason its something different.

• For the first part, you are totally correct. – Claude Leibovici Aug 29 '18 at 5:54
• For the first part "they" are correct, see my answer. For the second part you are right so far, but you are not finished because your answer should be in terms of $x$. – David Aug 29 '18 at 6:03

Starting at $(*)$ - I added a label to your question for the sake of explanation - you multiplied the LHS by $x(x^2)(x+7)$. But you multiplied the RHS by $x^2(x+7)$ which is different.

It would be best to multiply both sides by $(x^2)(x+7)$. You don't need the extra $x$ since it's already a factor of this.

Somehow they got $Ax(x+2)+B(x+2)+C(x^2)=x+7$

That is not even possible unless you factor out an x on the left hand side and divide both sides by x which would give you $\frac{x+7}{x}$ on the right hand side. So either their work is incorrect or I am missing something.

Yes, you're missingthe notion of least common denominator: $$\frac{A}{x}+\frac{B}{x^2}+\frac{c}{x+2}=\frac{x+7}{x^2(x+2)}$$ Here it is $x^2(x+2)$, and $$\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+2}= \frac{Ax(x+2)}{x^2(x+2)}+\frac{B(x+2)}{x^2(x+2)}+\frac{Cx^2}{x^2(x+2)},$$ so that multiplying by the denominator, you obtain indeed $$Ax(x+2)+B(x+2)+Cx^2=x+7.$$

• Now setting $x=0$, you get $2B=7$, so $B=\frac72$.

• Next $x=-2$ yields $4C=5$, or $C=\frac54$.

• Finally the coefficient of $x^2$ in the l.h.s. must be $0$, so $A+C=0$, and $A=-\frac 54$.