# What substitution to use in this type of integral?

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1. Mar 29, 2016

### gruba

1. The problem statement, all variables and given/known data
Find the integral $\int \frac{3x+1}{(x^2-x-6)\sqrt{3x^2+4x-7}}\mathrm dx$

2. The attempt at a solution
I have tried the types of substitutions of irrational functions, and Euler substitutions.
However, it seems that nothing simplifies this integral.

What substitution is useful for this type of integrals?

2. Mar 29, 2016

On this one, I would try completing the square of the polynomial inside the square root sign in the denominator. Once you do that, possibly a linear translation to get this part as sqrt(z^2-1). Then let $z=sec(\theta)$. I don't know that this will work, but it might be worth a try. This one looks complex in any case.

3. Apr 1, 2016

### vela

Staff Emeritus
Try starting with $u=3x+1$. That turns out to make the inside of the radical relatively nice.

4. Apr 1, 2016

Can you double-check this one and/or explain further? This problem stumped me as well. I tried my solution and it doesn't seem to work either because mine results in an integration of a couple of terms of the form $du/(A+B\sec(u))$ which I don't have an answer for.

5. Apr 1, 2016

### micromass

Staff Emeritus
https://en.wikipedia.org/wiki/Tangent_half-angle_substitution

6. Apr 1, 2016

### vela

Staff Emeritus
I didn't actually work it out. I just tried the "obvious" substitutions based on the terms in the integrand and noticed that one made the stuff inside the radical look nice.

Anyway, I futzed with it a bit more, and it looks like you can get it down to an integrand (I left out some overall constant factors) which looks like
$$\frac{5-\cos\theta}{(5-11\cos\theta)(5+4\cos\theta)}\,d\theta,$$ at which point you can proceed with the technique micromass pointed to.

7. Apr 1, 2016

I worked through it a little further using the $z=sec(\theta)$ mentioned in my first response. With some partial fractions, I got it to the form of $du/(A+B\sec(\theta)$. (Two terms of this type.) (I don't get the cosines that vela has.) The $z=tan(x/2)$ substitution micromass suggested works on these terms to get something that looks like $du/(1+u^2)$ which integrates to inverse tangent. (I had previously seen that substitution many years ago when we used the Thomas Calculus text.) Thank you both for your inputs. The OP doesn't seem to be responding, but it is an interesting calculation. I'm currently retired, and simply doing it as a hobby... A follow-on: I see to get vela's cosine terms from terms with secants, all that is necessary is to multiply numerator and denominator by $cos(\theta)$....