Integration using a given substitution.

**Furyan** · February 11th 2013, 12:34 PM

Hello,

I have to solve the following using the given substitution:

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

Using the substitution:

$u^2 = x - 1$

Which differentiating implicitly gives me:

$\dfrac{dx}{du} = 2u$

Substituting into to

$\int\dfrac{1}{1 + \sqrt{x - 1}} \dfrac{dx}{du}du$

I get

$\int\dfrac{1}{1 + u}(2u) du$

$2\int\dfrac{u}{1 + u} du$

Would someone please tell me if this is correct so far and if so how to proceed from here.

Thank you.

**EliteAndoy** · February 11th 2013, 12:43 PM

I think what you came up is correct assuming all the given is correct. To proceed from there, say that n=1+u, so that dn=du. Since n=1+u, we can say that u=n-1. You should come up with integral of (n-1)/n dn, so that if you simplify, you get integral of 1-1/n dn. If you integrate that you get 2[n-ln(abs(n))] + c. Sub back then you get 2(1+u)-2ln(abs(1+u))+c. Sub back again, you get 2(1+sqrt(x-1))-2ln(abs(1+sqrt(x-1)))+c. Im not sure though if you would need that Plus or minus from the root. By the way, sorry for not using latex, I'm kinda on a rush.

**MathJack** · February 11th 2013, 12:43 PM

yes, you could subsitute k for 1 + u at this point and work back after integrating

**Furyan** · February 11th 2013, 01:27 PM

Hello,

Thank you both very much using the method you suggested I got:

$2(1 + \sqrt{x - 1}) - 2\ln\mid1 + \sqt{x-1}\mid + c$

Is anyone able to tell me if that is correct.

Thank you

**MathJack** · February 11th 2013, 01:37 PM

Made mistake on final reverse substitution

**Furyan** · February 11th 2013, 02:17 PM

Originally Posted by MathJack

Made mistake on final reverse substitution

Thanks I thought it seemed to go too well. Unless it's equating $u = \sqrt{x-1}$ , I can't see where I've gone wrong. Here's what I did.

Let $k = 1 + u$

so $u = k - 1$

$2\int\dfrac{k - 1}{k} dk$

$2\int\1 - \dfrac{1}{k}dk$

$2(k - \ln(k))$

$2k - 2\ln(k)$

$K = 1 + u$ and $u =\sqrt{x - 1}$

$2(1 + u) - 2\ln(1 + u) + c$

$2(1 + \sqrt{x - 1}) - 2\ln(1 + \sqrt{x - 1}) + c$

Thank you.

**Furyan** · February 12th 2013, 04:45 PM

Hello,

I can see now that if at this stage:

$2\int\1 - \dfrac{1}{k}dk$ , I had replaced $k$ with $u + 1$ to get.

$2\int\1 - \dfrac{1}{u + 1}du$ , then I would have got the correct answer.

$2(u - ln(u + 1) + c$

I see that $\dfrac{u}{u + 1} = 1 - \dfrac{1}{u + 1}$ , which I'll be remembering, so I think the above is the way to go, but I'd like to know why it didn't work by integrating first and then substituting back in, as suggested. Did I make a mistake or can it not be done that way. I'd really appreciate it if someone would clarify that for me. Thank you.

**Prove It** · February 12th 2013, 05:54 PM

Originally Posted by Furyan

Hello,

I have to solve the following using the given substitution:

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

Using the substitution:

$u^2 = x - 1$

Which differentiating implicitly gives me:

$\dfrac{dx}{du} = 2u$

Substituting into to

$\int\dfrac{1}{1 + \sqrt{x - 1}} \dfrac{dx}{du}du$

I get

$\int\dfrac{1}{1 + u}(2u) du$

$2\int\dfrac{u}{1 + u} du$

Would someone please tell me if this is correct so far and if so how to proceed from here.

Thank you.

Here's how I would do it:

$\displaystyle \begin{align*} \int{\frac{1}{1 + \sqrt{x - 1}}\,dx} &= \int{\frac{1 - \sqrt{x - 1}}{2 - x}\,dx} \end{align*}$

Now let $\displaystyle \begin{align*} u = x - 1 \implies du = dx \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{\frac{1 - \sqrt{x - 1}}{2 - x}\,dx} &= \int{\frac{1 - \sqrt{u}}{1 - u}\,du} \\ &= 2\int{\frac{\sqrt{u} \left( 1 - \sqrt{u} \right)}{2\sqrt{u} \left( 1 - u \right)}\,du} \end{align*}$

Now let $\displaystyle \begin{align*}v = \sqrt{u} \implies dv = \frac{1}{2\sqrt{u}}\,du \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} 2\int{\frac{\sqrt{u}\left( 1 - \sqrt{u} \right) }{2\sqrt{u}\left( 1 - u \right) }\,du} &= 2\int{\frac{v \left( 1 - v \right)}{1 - v^2}\,dv} \\ &= 2\int{\frac{v}{1 + v}\,dv} \\ &= 2\int{ 1 - \frac{1}{1 + v} \,dv } \\ &= 2\left( v - \ln{\left| 1 + v \right|}\right) + C \\ &= 2 \left( \sqrt{u} - \ln{\left| 1 + \sqrt{u} \right|} \right) + C \\ &= 2\left( \sqrt{x - 1} - \ln{\left| 1 + \sqrt{ x - 1 } \right|} \right) + C \end{align*}$

**EliteAndoy** · February 12th 2013, 08:09 PM

I'm not entirely sure about this, but if you expand your final answer, u will get $2+2\sqrt{x-1}-2ln(1+\sqrt{x-1})+c$ , and you can also say that expression is equal to: $2\sqrt{x-1}-2ln(1+\sqrt{x-1})+(c+2)$ since if you take the derivative of that integral to revert back to your orginal function, that 2 would be treated as a constant. By that then, we can say that $\int\dfrac{1}{1+\sqrt{x-1}}dx=2\sqrt{x-1}-2ln(1+\sqrt{x-1})+c$ which is the correct answer. Don't count me on this one though.

**EliteAndoy** · February 12th 2013, 08:15 PM

Hi Prove! I think your way is the best way to do it. I actually would rationalize the function first before I proceed to integrate it but I think his original problem asks us to use that certain u substitution.

**Furyan** · February 13th 2013, 01:12 PM

Hello ProveIt,

Thank you very much, once again, for your efforts. They are very much appreciated. This weeks question were about using a given substitution, but I'm sure next week we will have to figure out for ourselves what substitution to use, so it's very helpful to see how best to solve the question. It's also good to know that I can use:

$\int\dfrac{u}{u + 1} = \int1 - \dfrac{1}{u + 1}$ , knowing this step would have made it much easier, but then if I hadn't struggled with it so much I probably wouldn't understand it.

Thanks again.

Hello EliteAndoy,

Thank you too for your efforts. They are also much appreciated. Looking at that '2' I had briefly thought, 'maybe that's just part of the constant term'. Now you've pointed it out it does make sense that it would be.

Thank you both very much for your help

**EliteAndoy** · February 13th 2013, 02:04 PM

No problem man. I'd be more than happy to help you again with integration problems like this

.

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