Hi Guys,

Thank you

2. use the substitution x = e^t and then perform integration by parts.

3. Originally Posted by aeonboy
Hi Guys,

Thank you
I like Peritus' substitution

alternatively:

we could split the integral up and find,

$\displaystyle \int_1^e \frac {\ln x}{\sqrt{x}}~dx + \int_1^e \frac 1{\sqrt{x}}~dx$

the last integral can be done using the power rule, for the first, i see two ways of approaching it.

the easier of the two ways is to make the substitution $\displaystyle u = \sqrt{x}$, and continue by integration by parts, which will be easy (you may even know the integral of ln x by heart, so integration by parts won't be necessary)

the second way is to multiply the first integral by $\displaystyle \frac {\sqrt{x}}{\sqrt{x}}$ and then use the substitution $\displaystyle u = \ln x$. again you would continue by parts, but this one would be a little harder i think

4. Originally Posted by Peritus
use the substitution x = e^t and then perform integration by parts.
Hi Peritus,

Thanks for your help but i am not sure how to use the substitution and then perform integration by parts.

Could anyone guide me with the steps?

Thank you

5. Originally Posted by aeonboy
Hi Peritus,

Thanks for your help but i am not sure how to use the substitution and then perform integration by parts.

Could anyone guide me with the steps?

Thank you
using Jhevon's solution, we are left to find the $\displaystyle \int \frac{ln x}{\sqrt x} \, dx$

if we set, $\displaystyle u = \sqrt x \implies u^2 = x$ (we can assume this is true since x is positive, otherwise, we need the absolute value of x..)

now, $\displaystyle du = \frac{1}{2\sqrt x} \, dx \implies 2du = \frac{1}{\sqrt x} \, dx$

so we have $\displaystyle \int {2 ln \, u^2 \, du} = 4\int {ln \, u \, du}$

and like what he said, you should know the integral of ln x by heart.. Ü

6. Consider $\displaystyle f(x)=x\ln x.$ Take its derivative $\displaystyle f'(x)=\ln x+1.$

Integrate $\displaystyle f(x)+k=\int\ln x\,dx+x.$

And we happily get $\displaystyle \int\ln x\,dx=x\ln x-x+k.$

7. Originally Posted by Krizalid
Consider $\displaystyle f(x)=x\ln x.$ Take its derivative $\displaystyle f'(x)=\ln x+1.$

Integrate $\displaystyle f(x)+k=\int\ln x\,dx+x.$

And we happily get $\displaystyle \int\ln x\,dx=x\ln x-x+k.$
hehe. i like this approach, when you use a derivative to find the integral. it is a neat concept. sometimes it is hard to come up with the function to differentiate, but when you can come up with it, it seems all too obvious. (in fact, the method of integration by parts was developed using this approach. what they did was to undo the product rule)

for the benefit of the class, i shall show how we find the integral of ln(x) using the by parts method.

$\displaystyle \int ln x~dx$

the trick here is to note that $\displaystyle \ln x = 1 \cdot \ln x$, obvious enough

Thus, if we take $\displaystyle u = \ln x \implies u' = \frac 1x$ and $\displaystyle v' = 1 \implies v = x$, we have:

$\displaystyle \int uv' = uv - \int v'u$

$\displaystyle \Rightarrow \int \ln x = x \ln x - \int x \cdot \frac 1x~dx$

$\displaystyle = x \ln x - \int ~dx$

$\displaystyle = x \ln x - x + C$

An alternative method to find the integral of $\displaystyle \ln x$ is the following. (i actually just thought of this now, i've never seen this done before, but the technique is a very familiar one)

$\displaystyle \int \ln x~dx = \int \frac xx \ln x~dx$

By substitution, let $\displaystyle u = \ln x \implies x = e^u$

$\displaystyle \Rightarrow du = \frac 1x~dx$

Thus, our integral becomes

$\displaystyle \int u e^u~du = ue^u - \int e^u~du$ (by parts)

$\displaystyle = u e^u - e^u + C$

back-substitute:

$\displaystyle = \ln x \cdot e^{\ln x} - e^{\ln x} + C$

$\displaystyle = x \ln x - x + C$

as desired

of course, this way is a bit more work than the last, but it's nice to know several ways to skin a cat...if skinning a cat is your thing

8. I have managed to solve it already, thanks to all of you for your help