Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\displaystyle\int \left(\frac {x-1}{3-x}\right)^\frac{1}{2}\,\rm dx$

I am stuck on this part:

Let $u=\dfrac{x-1}{3-x}~\longrightarrow$ $~~\rm du=\dfrac {2}{(x-3)^2}\,\rm dx,$ which can be represented as

$\rm du=\dfrac{1}{3-x} - \dfrac{1-x}{(3-x)^2}\,\rm dx$

I cannot "see" how to get to this $2$ $\displaystyle\int \: \frac{(u)^\frac{1}{2}}{(u+1)^2} \:\rm dx$

after this part I know how to solve it; I just wish someone would show me "step by step" this part It seems it involves some sort of "leap" of thought; or is there a systematic way doing this using basic algebra?

Thanks.

share|improve this question
add comment

6 Answers 6

First note $$ du=\frac{2}{(x-3)^2}dx $$ which implies $$ \frac{(x-3)^2}{2}du=dx. $$ Also, $$ u+1=\frac{x-1}{3-x}+\frac{3-x}{3-x}=\frac{2}{3-x} $$ which implies $$ (u+1)^2=\left(\frac{2}{3-x}\right)^2=\frac{4}{(x-3)^2}. $$

Putting it together you have $$ \int \left(\frac {x-1}{3-x}\right)^\frac{1}{2} dx = \int u^{1/2}\cdot\frac{(x-3)^2}{2}du = 2\int u^{1/2}\cdot\frac{(x-3)^2}{4}du = 2\int \frac{u^{1/2}}{(u+1)^2}du. $$

share|improve this answer
add comment

You probably reached this point after inserting u:

$\int u^\frac{1}{2} \frac{(x-3)^2}{2}du$

The leap of thought you would need here is to realize that the remaining x must be substituted by u and so what you want to find is a function which in terms of u can replace x, in this case we can go for $(x-3)$.

$x-3 = f(u)$

A trial and error approach would be to inspect $u$ and see that (as you realized) $u$ can be written as: $u = -1 + \frac{2}{3-x}$, here we already see the term $3-x$ so we just need to manipulate the equation to let $x-3$ be on one side. step by step:

$u+1 = \frac{2}{3-x}$

$\frac{2}{u+1} = 3-x$

$-\frac{2}{u+1} = x-3$

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

Now we can substitute $(x-3)$ with $-\frac{2}{u+1}$ in the integral to get $2$ $\int \frac{(u)^\frac{1}{2}}{(u+1)^2}dx$

share|improve this answer
    
why the downvote? –  j-a Jan 24 '12 at 17:46
add comment

Notice the $u+1$ in the denominator. What is $u+1$ in terms of $x$? In particular, what is $\frac{1}{(u+1)^2}$?

share|improve this answer
add comment

You can write the fraction as $\frac{(x-1)^{1/2}}{(3-x)^{1/2}}$. Then put $3-x =t$. So you have $dt= -dx$ and $x-1 =3-t-1=2-t$. So you have to now evaluate the integral $$\int \sqrt{ \frac{2-t}{t}}\ -\rm{dt}$$ which can be evaluated by sing the subsitution $t = 2 \cos^{2}{v}$.

share|improve this answer
add comment

You only came unstuck at the point where you needed to set $\text{d}x$ equal to $\frac{(3-x)^2}{2} \text{d}u,$ an equality you already had. So your problem reduces to expressing $3-x$ in terms of $u.$

From your expression for $u$ you have $u(3-x)=x-1,$ and adding $3-x$ to both sides gives $(3-x)(u+1)=2$ and hence $3-x=2/(1+u).$

The sure, but slightly longer approach, is to solve $u(3-x)=x-1$ for $x$ and subtract from $3.$

share|improve this answer
add comment

The integrand is so seductively pointing to the use of complex variables which reduces it to a trivial problem. Beware we have a pole within a branch cut!

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.