Hard integration by parts question,

[limelightdevo]

Homework Statement

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

Homework Equations

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

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)

Please help

Thankssss

 

[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.

 

[vela]

You might want to learn how to write equations here.

https://www.physicsforums.com/showthread.php?t=546968

 

[christoff]

[itex]t^2-9=(t-3)(t+3)[/itex]. Partial fractions, and you're done.

 

[vela]

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

$$\int_4^6 \frac{1}{t^2-9}\,dt$$

so I multiplied the top and bottom by square root of 9.

Also known as 3.

which got me square root of (t/3)^2 - 1 in the denominator.

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.

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)

Please help

 

[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.