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

MATH101 April 2011

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Question 04 (a)
Evaluate the integral (hint: put $u={\sqrt {x}}$). $\int {\frac {\sqrt {x}}{x-1}}dx$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
If $u={\sqrt {x}}$, then $du={\frac {1}{2{\sqrt {x}}}}dx$.

Hint 2
Use partial fractions.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 ${\begin{aligned}\int {\frac {\sqrt {x}}{x-1}}dx&=2u+\int {\frac {2}{(u-1)(u+1)}}du\\&=2u+\int {\frac {1}{u-1}}du-\int {\frac {1}{u+1}}du\\&=2u+\ln |u-1|-\ln |u+1|+C\\&=2{\sqrt {x}}+\ln |{\sqrt {x}}-1|-\ln |{\sqrt {x}}+1|+C.\end{aligned}}$

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