# Hard integration by parts question, need help

1. Jun 23, 2012

### limelightdevo

1. The problem statement, all variables and given/known data

Integral, from 4 to 6, 1/(t^2-9) dt

2. Relevant equations

please use my approach to solve it, like with cosh and whatnot

3. The attempt at a solution

Integral, from 4 to 6, 1/(t^2-9) dt

so I multiplied the top and bottom by square root of 9.
which got me square root of (t/3)^2 - 1 in the denominator.
so I got integrated that into arccosh (t/3) - arccosh (t/3) from 4 to 6
after some solving, it came down to square root of 9 times ((ln(2)+squareroot3))-(ln(4/3)+squareroo… And that is where I got stuck. If I distribute in the square root 9,it should be (3 ln(2) + square root 3) - (3 ln (4/3) + square root (7/9)), which results in (ln 8 + 3*square root 3) - (ln (4/3)^3 + 3*square root (7)). But when I distribute the square root 9 in and when it multiplies with ln(2), for example, it should be ln 2^3 and not ln 2*3 based on the multiplication rule of natural log. But by doing the wrong way, it perfectly leads to the answer, which is ln(6+3*square root 3)/(3+square root 7)

Thankssss

2. Jun 24, 2012

### NewtonianAlch

I'm a bit confused as to the approach you're taking. If you split that given fraction to get partial fractions, you can integrate each term easily, for which will you get log functions, substitute the values and then you're done.

3. Jun 24, 2012

### vela

Staff Emeritus
4. Jun 25, 2012

### christoff

$t^2-9=(t-3)(t+3)$. Partial fractions, and you're done.

Last edited: Jun 25, 2012
5. Jun 25, 2012

### vela

Staff Emeritus
$$\int_4^6 \frac{1}{t^2-9}\,dt$$
Also known as 3.
What you've said so far doesn't make sense. This is what you said you did:
$$\int_4^6 \frac{1}{t^2-9}\,dt = \int_4^6 \frac{3}{3(t^2-9)}\,dt \ne \int_4^6 \frac{1}{\sqrt{(t/3)^2-1}}\,dt$$ The rest of what you did follows from this mistake, so it's all wrong. What I think you meant was that you factored 9 out of the denominator:
$$\int_4^6 \frac{1}{t^2-9}\,dt = \int_4^6 \frac{1}{9[(t/3)^2-1]}\,dt$$ Still, the square root wouldn't appear out of nowhere. Or perhaps you mistyped the original problem here, and what you did was
$$\int_4^6 \frac{1}{\sqrt{t^2-9}}\,dt = \int_4^6 \frac{1}{\sqrt{9[(t/3)^2-1]}}\,dt = \int_4^6 \frac{1}{3\sqrt{(t/3)^2-1}}\,dt$$ because that would be okay so far.

6. Jun 25, 2012

### ougoah

Think you forgot to put a square root in your question, or you wouldn't have got the answer you got. I used a sec substitution and got $\displaystyle\ln\left(\frac{6+3\sqrt{3}}{4+\sqrt{7}}\right)$, which is slightly different to your answer.

Last edited: Jun 25, 2012