Integral using given substitute...

**flashylightsmeow** · April 3rd 2011, 09:58 AM

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

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

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

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

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.

**topsquark** · April 3rd 2011, 10:15 AM

Originally Posted by flashylightsmeow

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

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

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

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

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.

$\displaystyle \int \frac{\sqrt{t - 1}}{t - 2}~dt = \int \frac{2u^2}{u^2 - 1}~du$ which can be evaluated using a number of different methods. What answer are you getting?

-Dan

**Sambit** · April 3rd 2011, 10:19 AM

Originally Posted by flashylightsmeow

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

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

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

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

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

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

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

**Sambit** · April 3rd 2011, 10:21 AM

Originally Posted by flashylightsmeow

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

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

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

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

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

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

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

EDIT: ooops....sorry

**flashylightsmeow** · April 3rd 2011, 10:49 AM

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

**topsquark** · April 3rd 2011, 11:06 AM

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?

$\displaystyle \frac{1}{u^2 - 1} = \frac{A}{u + 1} + \frac{B}{u - 1}$

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

What are A and B?

-Dan

**flashylightsmeow** · April 3rd 2011, 11:34 AM

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

**topsquark** · April 3rd 2011, 12:31 PM

Originally Posted by flashylightsmeow

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

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

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

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

Can you take it from here?

-Dan

**flashylightsmeow** · April 3rd 2011, 01:05 PM

Got it! Thank you very much.

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