1
$\begingroup$

I have the following integral: $$y=\int \frac{1}{1-2\sqrt{x}} \, dx$$

I first got $u=2\sqrt{x}$ which gives us $x=\frac{u^2}{4}$. Plugging this in I got: $$y=\int \frac{1}{1-2\sqrt{\frac{u^2}{4}}} \, du=\int \frac{1}{1-2(\frac{u}{2})} \, du=\int \frac{1}{1-u} \, du$$

After this I am not sure where to go from there. Even though there might be easier methods I'd prefer to stick with this method so please relate your responses to my correct or incorrect work shown above.

$\endgroup$
5
  • $\begingroup$ How did you substitute out $dx$? $\endgroup$ Jul 30, 2020 at 19:38
  • 1
    $\begingroup$ If $u = 2\sqrt{x}$ then $dx = \frac{u}{2}du$ and you need to take care of it, too getting $y = \int \frac{u}{2(1-u)}du$. Then maybe another substitution of type $t = 1-u$, $dt = -du$ can help you to finish? $\endgroup$ Jul 30, 2020 at 19:44
  • $\begingroup$ What is the derivative of $-\ln(1-u)$? $\endgroup$
    – LL 3.14
    Jul 30, 2020 at 23:22
  • $\begingroup$ I did the other substitution and got $-\int \frac{dt}{t}=-ln(t)=-ln(1-u)=-ln(1-2sqrt{x})$. However when I take the deriative of this I do not get what I started with. What am I doing wrong @DominikKutek $\endgroup$
    – user710744
    Jul 31, 2020 at 19:23
  • $\begingroup$ @LL3.14 same to you^ $\endgroup$
    – user710744
    Jul 31, 2020 at 19:24

3 Answers 3

0
$\begingroup$

We have $$ \int \frac{1}{1-2\sqrt{x}}dx$$ Substitute $u = 2\sqrt{x}$ so that $x = \frac{u^2}{4}$ and $dx = \frac{u}{2}du$. We get $$ \int \frac{u}{2(1-u)}du $$ Now, it can be done from here, but since you're learning, make another substitution which might help you seeing "typical" integral here.

Let $t = 1-u$ so that $u= 1-t$ and $du = -dt$. We get $$ \int \frac{1-t}{2t}(-dt) = -\frac{1}{2}\int \frac{1}{t} - \frac{t}{t} dt = -\frac{1}{2} \Big ( \int \frac{1}{t} dt - \int 1 dt \Big) $$

You should know those integrals. Preciselly: $$ \int \frac{1}{t} dt = \ln|t| + C_1 $$ and $$ \int 1 dt = t + C_2$$

So that taking $C=-\frac{1}{2}(C_1 + C_2) \in \mathbb R$ we have: $$ \int \frac{1}{1-2\sqrt{x}}dx = -\frac{1}{2} \big ( \ln|t| - t \big) + C $$ and now go back to variable $x$:

$$\int \frac{1}{1-2\sqrt{x}}dx= -\frac{1}{2}\ln|1-u| + \frac{1}{2}(1-u) + C = -\frac{1}{2}\ln|1-2\sqrt{x}| + \frac{1}{2}(1-2\sqrt{x}) + C $$

$\endgroup$
0
$\begingroup$

The substitution step that you made is incorrekt. Regarding the integral: $$\int{\frac{dx}{1-2\sqrt{x}}}$$ Let $x(\phi) = \phi^2/4$, then $\frac{dx}{d\phi}=\frac{\phi}{2}$, now substitute it: $$\frac{1}{2}\int{\frac{\phi}{1-\phi}d\phi}$$ To make it nicer to work with, you can rearrange it as follows: $$ \begin{align} \frac{1}{2}\int{\frac{1}{1-\phi}-1\ d\phi} &= -\frac{1}{2}\left(\log{|1-\phi|}+\phi\right)+K \\ &= -\frac{1}{2}\left(\log{|1-2\sqrt{x}|}+2\sqrt{x}\right)+K \end{align} $$ Where $K$ is an arbitrary constant.

$\endgroup$
0
$\begingroup$

here is my nice answer is Let $u= 1-2\sqrt x$, $du=-dx/\sqrt x$, $(u-1)du/2=dx$,

$\int \dfrac{dx}{1-2\sqrt x}$

$=\int \dfrac{(u-1)/2\ du}{u}$

$=\dfrac12\int (1-\frac1u)du$

$=\dfrac12 (u-\ln|u|)+c$

put $u=1-2\sqrt x$

$=-\dfrac12 (2\sqrt x+\ln|1-2\sqrt x|)+c$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .