# Thread: indefinite integral

1. ## indefinite integral

Hi, I've attached an indefinite integral problem I thought I did correctly, but the answer turned out to be wrong. If anyone could point out my error I would greatly appreciate it! Thanks

2. ## Re: indefinite integral

I would do it a different way...

\displaystyle \begin{align*} \int{\frac{\sqrt{x-1}}{x}\,dx} &= \int{\frac{(\sqrt{x-1})^2}{x\sqrt{x - 1}}\,dx} \\ &= 2\int{\frac{x - 1}{x}\cdot \frac{1}{2\sqrt{x - 1}}\,dx} \end{align*}

Now make the substitution $\displaystyle u = \sqrt{x - 1} \implies du = \frac{1}{2\sqrt{x - 1}}\,dx$ and also $\displaystyle x = u^2 + 1$, then the integral becomes

\displaystyle \begin{align*} 2\int{\frac{x - 1}{x} \cdot \frac{1}{2\sqrt{x - 1}}\,dx} &= 2\int{\frac{u^2}{u^2 + 1}\,du} \\ &= 2\int{1 - \frac{1}{u^2 + 1}\,du} \\ &= 2\left(u - \arctan{u}\right) + C \\ &= 2\left[\sqrt{x-1} - \arctan{\left(\sqrt{x-1}\right)}\right] + C \end{align*}

3. ## Re: indefinite integral

Your first substitution is not correct, you just said: u=sqrt(1-x) and substitued without calculating correctly du in function of dx.

EDIT:
Solution with your substitution(s):
$\int \frac{\sqrt{x-1}}{x}dx=\int \frac{\sqrt{x-1}}{(x-1)+1}dx$
Let $\sqrt{x-1}=t$ so $\frac{dx}{2\sqrt{x-1}} \Leftrightarrow dx=2tdt$
So the integral becomes:
$\int \frac{2t^2dt}{t^2+1}dt=2\int \frac{(t^2+1)-1}{t^2+1}dt=2\int \frac{t^2+1}{t^2+1}dt - 2\int \frac{dt}{t^2+1}=2t-2\arctan(t) + C = 2\sqrt{x-1}-2\arctan\left(\sqrt{x-1}\right)+C$

4. ## Re: indefinite integral

Write $\int{\frac{\sqrt{x-1}}{x}\,dx}=\int{\frac{x-1}{x\sqrt{x-1}}\,dx}=\int{\frac{dx}{\sqrt{x-1}}}-\int{\frac{dx}{x\sqrt{x-1}}},$ in the last integral put $x=\frac1{t^2},$ so the original integral equals $2\left( \sqrt{x-1}+\arcsin\frac1{\sqrt x} \right)+k.$

5. ## Re: indefinite integral

Originally Posted by Krizalid
Write $\int{\frac{\sqrt{x-1}}{x}\,dx}=\int{\frac{x-1}{x\sqrt{x-1}}\,dx}=\int{\frac{dx}{\sqrt{x-1}}}-\int{\frac{dx}{x\sqrt{x-1}}},$ in the last integral put $x=\frac1{t^2},$ so the original integral equals $2\left( \sqrt{x-1}+\arcsin\frac1{\sqrt x} \right)+k.$
I think you've made a mistake, if you let $\displaystyle x = t^{-2}$ then $\displaystyle dx = -2t^{-3}\,dt$, or $\displaystyle dt = -\frac{1}{2}t^3\,dx$... I'm not sure how you can convert $\displaystyle \int{\frac{dx}{x\sqrt{x-1}}}$ to $\displaystyle \int{\frac{dt}{\sqrt{t^2 - 1}}}$...

6. ## Re: indefinite integral

Try it again.

7. ## Re: indefinite integral

Originally Posted by Krizalid
Try it again.
Well I realise now that if $\displaystyle x = t^{-2}$ and $\displaystyle dt = -\frac{1}{2}t^3\,dx$ then $\displaystyle dt = -\frac{1}{2}x^6\,dx$... Still don't see how you're going to use this substitution to your advantage...

8. ## Re: indefinite integral

You don't see it?

$\int{\frac{dx}{x\sqrt{x-1}}}=-2\int{\frac{1}{{{t}^{3}}}\cdot \frac{1}{\frac{1}{{{t}^{2}}}\sqrt{\frac{1}{{{t}^{2 }}}-1}}\,dt}=-2\int{\frac{{{t}^{3}}}{{{t}^{3}}\sqrt{1-{{t}^{2}}}}\,dt}=-2\int{\frac{dt}{\sqrt{1-{{t}^{2}}}}}.$

9. ## Re: indefinite integral

Originally Posted by Krizalid
You don't see it?

$\int{\frac{dx}{x\sqrt{x-1}}}=-2\int{\frac{1}{{{t}^{3}}}\cdot \frac{1}{\frac{1}{{{t}^{2}}}\sqrt{\frac{1}{{{t}^{2 }}}-1}}\,dt}=-2\int{\frac{{{t}^{3}}}{{{t}^{3}}\sqrt{1-{{t}^{2}}}}\,dt}=-2\int{\frac{dt}{\sqrt{1-{{t}^{2}}}}}.$
OK I see it now, thanks...

10. ## Re: indefinite integral

We coulda done the remaining integral by putting $t=\sqrt{x-1},$ which gives an arctangent.

11. ## Re: indefinite integral

Originally Posted by Krizalid
We coulda done the remaining integral by putting $t=\sqrt{x-1},$ which gives an arctangent.
Which I did...

12. ## Re: indefinite integral

But on my remaining integral it leads a direct integral, in your remaining integral you had to do a long division, so it's one less step.