1. ## integration by substitution

Hi there,

I've been given the problem attached and its causing me lots of problems. When I perform the substitution I don't know how to "correct" du so that the new integral has a denominator that is equivalent to t-2.

2. ## Re: integration by substitution

With the given substitution:

$\displaystyle u=\sqrt{t-1}\,\therefore\,du=\frac{1}{2u}\,dt\,\therefore\,d t=2u\,du$

If we square the substitution:

$\displaystyle u^2=t-1\,\therefore\,t-2=u^2-1$

Can you proceed?

3. ## Re: integration by substitution

Originally Posted by flashylightsmeow
Hi there,

I've been given the problem attached and its causing me lots of problems. When I perform the substitution I don't know how to "correct" du so that the new integral has a denominator that is equivalent to t-2.
\displaystyle \displaystyle \begin{align*} \int{ \frac{ \sqrt{ t - 1 } }{ t - 2 } \, dt } &= 2 \int{ \frac{ t - 1 }{ 2 \, \sqrt{ t - 1} \left( t - 2 \right) } \, dt } \end{align*}
Now make the substitution \displaystyle \displaystyle \begin{align*} u = \sqrt{ t - 1} \implies du = \frac{1}{2 \, \sqrt{t - 1} } \, dt \end{align*} and the integral becomes
\displaystyle \displaystyle \begin{align*} 2 \int{ \frac{ u^2 }{ u^2 - 1 } \, du } &= 2 \int{ 1 + \frac{1}{u^2 - 1} \, du} \end{align*}