# Thread: Integral using given substitute...

1. ## Integral using given substitute...

Hi there, I'm having some difficulty with the following problem:

Integrate: $\displaystyle \int \frac{\sqrt{t-1}}{t-2}dt$ using u=$\displaystyle \sqrt{t-1}$

I've found that the denominator = $\displaystyle u^2-1$ and

$\displaystyle \frac{1}{2(t-1)^\frac{1}{2}}dt = \frac{1}{2u}$ and therefore $\displaystyle dt=2u\ du$

But am not sure if I'm on the right track and where to go from there, because any progress I've made doesn't give me the answer in the back of the book.

2. Originally Posted by flashylightsmeow
Hi there, I'm having some difficulty with the following problem:

Integrate: $\displaystyle \int \frac{\sqrt{t-1}}{t-2}dt$ using u=$\displaystyle \sqrt{t-1}$

I've found that the denominator = $\displaystyle u^2-1$ and

$\displaystyle \frac{1}{2(t-1)^\frac{1}{2}}dt = \frac{1}{2u}$ and therefore $\displaystyle dt=2u\ du$

But am not sure if I'm on the right track and where to go from there, because any progress I've made doesn't give me the answer in the back of the book.
$\displaystyle \displaystyle \int \frac{\sqrt{t - 1}}{t - 2}~dt = \int \frac{2u^2}{u^2 - 1}~du$ which can be evaluated using a number of different methods. What answer are you getting?

-Dan

3. Originally Posted by flashylightsmeow
Integrate: $\displaystyle \int \frac{\sqrt{t-1}}{t-2}dt$ using u=$\displaystyle \sqrt{t-1}$

I've found that the denominator = $\displaystyle u^2-1$ and

$\displaystyle \frac{1}{2(t-1)^\frac{1}{2}}dt = \frac{1}{2u}$ and therefore $\displaystyle dt=2u\ du$
Correct till this. Then $\displaystyle \int\frac{u}{u^2-1}(2u.du) = \int 2\frac{u^2}{u^2-1}du$

= $\displaystyle 2\int (1+\frac{1}{u^2-1})$

= $\displaystyle 2(u+\frac{1}{2}\log\frac{1-u}{1+u})$ + constant

Substitute back $\displaystyle u=\sqrt{t-1}$.

4. Originally Posted by flashylightsmeow
Integrate: $\displaystyle \int \frac{\sqrt{t-1}}{t-2}dt$ using u=$\displaystyle \sqrt{t-1}$

I've found that the denominator = $\displaystyle u^2-1$ and

$\displaystyle \frac{1}{2(t-1)^\frac{1}{2}}dt = \frac{1}{2u}$ and therefore $\displaystyle dt=2u\ du$
Correct till this. Then $\displaystyle \int\frac{u}{u^2-1}(2u.du) = \int 2\frac{u^2}{u^2-1}du$

= $\displaystyle 2\int (1+\frac{1}{u^2-1})$

= $\displaystyle 2(u+\frac{1}{2}\log\frac{1-u}{1+u})$ + constant

Substitute back $\displaystyle u=\sqrt{t-1}$.

EDIT: ooops....sorry

5. Ah ok, I got to this bit, and took the 2 outside the integral, but how do you split the integrand into partial fractions?

6. Originally Posted by flashylightsmeow
Ah ok, I got to this bit, and took the 2 outside the integral, but how do you split the integrand into partial fractions?
$\displaystyle \displaystyle \frac{1}{u^2 - 1} = \frac{A}{u + 1} + \frac{B}{u - 1}$

Thus
$\displaystyle A(u - 1) + B(u + 1) = 1 \implies A + B = 0 \text{ and } -A + B = 1$

What are A and B?

-Dan

7. so A=-1/2 and B=1/2 so $\displaystyle \int \frac{1}{2(u-1)}+\frac{1}{2(u+1)}$ but how do I get from this to Sambit's integral above?

8. Originally Posted by flashylightsmeow
so A=-1/2 and B=1/2 so $\displaystyle \int \frac{1}{2(u-1)}+\frac{1}{2(u+1)}$ but how do I get from this to Sambit's integral above?
$\displaystyle \displaystyle \int\frac{2u^2}{u^2-1}~du = 2 \int \left ( 1 + \frac{1}{u^2-1} \right )~du$

$\displaystyle \displaystyle = 2\int du + 2 \int \frac{1}{u^2-1}du = 2\int du + 2 \int \left ( \frac{-1/2}{u + 1} + \frac{1/2}{u - 1} \right ) ~du$

$\displaystyle \displaystyle = 2\int du - \frac{2}{2} \int \frac{1}{u + 1} + \frac{2}{2} \int \frac{1}{u - 1}~du$

Can you take it from here?

-Dan

9. Got it! Thank you very much.