Science:Math Exam Resources/Courses/MATH101/April 2011/Question 04 (a)/Solution 1

Set the following change of variable

$u={\sqrt {x}}$

then we have

$\displaystyle {x=u^{2}}$

and so

$\displaystyle {dx=2u\,du}.$

Making this substitution in the integral gives

$\int {\frac {\sqrt {x}}{x-1}}dx=\int {\frac {u}{u^{2}-1}}\,2u\,du=2\int {\frac {u^{2}}{u^{2}-1}}du.$

Now we use the method of partial fractions. We have

${\begin{aligned}\int {\frac {\sqrt {x}}{x-1}}dx&=2\int {\frac {u^{2}}{u^{2}-1}}du\\&=2\int {\frac {u^{2}-1+1}{u^{2}-1}}du\\&=2\int \left(1+{\frac {1}{u^{2}-1}}\right)du\\&=2u+\int {\frac {2}{u^{2}-1}}du\\&=u+\int {\frac {2}{(u-1)(u+1)}}du.\end{aligned}}$

We want to find numbers $A$ and $B$ such that

${\frac {2}{(u-1)(u+1)}}={\frac {A}{u-1}}+{\frac {B}{u+1}}.$

Multiplying both sides by $(u-1)(u+1)$ yields

$\displaystyle 2=Au+A+Bu-B.$

Equating coefficients of powers of $u$ on both sides gives

${\begin{cases}0=A+B\\2=A-B\end{cases}}$

From the first of these equations we see that $A=-B$. So the second equation becomes

$\displaystyle 2=2A.$

Hence $A=1$ and $B=-A=-1$. Therefore, we have

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

Using this to rewrite the last integral above, we find

${\begin{aligned}\int {\frac {\sqrt {x}}{x-1}}dx&=2u+\int {\frac {2}{(u-1)(u+1)}}du\\&=2u+\int {\frac {1}{u-1}}du-\int {\frac {1}{u+1}}du\\&=2u+\ln |u-1|-\ln |u+1|+C\\&=2{\sqrt {x}}+\ln |{\sqrt {x}}-1|-\ln |{\sqrt {x}}+1|+C.\end{aligned}}$