# Integration of Fractions

1. Aug 13, 2012

### hms.tech

What is the procedure to integrate this kind of a fraction, i am guessing some trigonometric
identity will apply but i am not quite sure how

[1/(a-bx^2)] dx

Any help is Appreciated

2. Aug 13, 2012

### DonAntonio

$$\frac{1}{a-bx^2}=\frac{1}{(\sqrt a-\sqrt bx)(\sqrt a+\sqrt bx)}$$
iff $\,a,b\geq 0\,$

DonAntonio

3. Aug 13, 2012

### mathman

The above works as long as a and b have the same sign - if both are negative, replace by minus absolute value. If they are of opposite sign, then you end up with something involving an arctan.

4. Aug 13, 2012

### johnqwertyful

Can't you just pull out a ${b}$, giving $\frac{1}{{b}(\frac{a}{{b}}-x^{2})}$

Then use a simple inverse sin formula?

5. Aug 14, 2012

### Millennial

You can, but that does not mean you have to. It depends on what solution is being asked from you, because for example you are asked to solve the problem with partial fractions and you try a sine substitution, you will have to break up your answer into two natural logarithms later on. For definite integrals though, this often is not a problem.

6. Aug 14, 2012

### DonAntonio

Inverse sine formula? For that one needs $\,(\sqrt{1-x^2})^{-1}\,$ and we have no square roots here.

DonAntonio

7. Aug 14, 2012

### Millennial

DonAntonio, it is still applicable. Let $\sin(t)=x$. Then we have that $\displaystyle \int \frac{dt}{\cos(t)}=\int \sec(t)dt$ which has a closed-form solution.

8. Aug 14, 2012

### johnqwertyful

Yeah, you need a square root for inverse sin, I knew that.
But still, just do a trig sub.

9. Aug 14, 2012

### DonAntonio

Yes...so? I was referring to other poster's idea of using something about the inverse sine, or $\,\arcsin\,$ , which I

can't see how it can appear here more or less "naturally". Of course, one can alwasy make up substitutions as to make

almost any function pop up, but this is not what I meant.

DonAntonio