1. ## Integrals!

Hi people!

I can't solve these ones, they're tough!

Solve $\displaystyle \int {\frac{{\sqrt {\sqrt {x^4 + 1} - x^2 } }} {{x^4 + 1}}\,dx}$

Evaluate $\displaystyle \int_0^1 {x\ln x\,dx}$

Thanks!

2. for the second one do integration by parts, you should get an answer of

$\displaystyle \frac{x^2 lnx}{2} - \frac{x^2}{4}$

3. The definition of "tough" always has eluded me.

Frequent Definition: I don't know how to do it.

Occasional Definition: I can do it, but it takes too long and requires too much effort.

Pratical Definition: It can be done, but who would want to?

Reasonable Definition: I could spend a lot of time on this, using some way that I have concocted, but even if it works, there simply MUST be an easier way.

1) Why have you been given these integrals? Are you REALLY expected to solve them in a closed form? Without a very specific hint, and with your missing he Integration by Parts on the second, it seems to me that the first is not reasonable. Do you have a specific hint?

2) Does one get to use numerical methods? Numbers are not evil. It's okay if that's what we have.

4. Originally Posted by Mess
Hi people!

I can't solve these ones, they're tough!

Solve $\displaystyle \int {\frac{{\sqrt {\sqrt {x^4 + 1} - x^2 } }} {{x^4 + 1}}\,dx}$

Evaluate $\displaystyle \int_0^1 {x\ln x\,dx}$

Thanks!
For the first one, try substituting $\displaystyle x^2 = tan(t)$.

You will end up with an integrand looking like:

$\displaystyle \frac{\sqrt{sec(t)-tan(t)}}{sec^2(t)}$

5. Here's a specific calculation in this case.

Originally Posted by Mess

Evaluate $\displaystyle \int_0^1 {x\ln x\,dx}$
$\displaystyle \ln x=\int_1^x\frac1u\,du.$ So,

$\displaystyle \int_0^1 {x\ln x\,dx} = - \int_0^1 {\int_x^1 {\frac{x} {u}\,du} \,dx} = - \int_0^1 {\underbrace {\int_0^u {x\,dx} }_{\frac{1} {2}u^2 }\frac{1} {u}\,du} .$

The rest is routine, and the answer is $\displaystyle -\frac14.$

Even I'd like to take a more general case:

Let $\displaystyle m,n>0.$

Define $\displaystyle \int_0^1x^m\ln^nx\,dx.$

Make substitution $\displaystyle u=-\ln x$ which can be written as $\displaystyle x=e^{-u}\implies dx=-e^{-u}\,du,$ the integral becomes

$\displaystyle \int_0^1 {x^m \ln ^n x\,dx} = ( - 1)^n \int_0^\infty {e^{ - (m + 1)u} u^n \,du} .$

Define another substitution according to

$\displaystyle \alpha = (m + 1)u \implies u = \frac{\alpha } {{m + 1}}\,\therefore \,du = \frac{1} {{m + 1}}\,d\alpha ,$

$\displaystyle \int_0^1 {x^m \ln ^n x\,dx} = \frac{{( - 1)^n }} {{(m + 1)^{n + 1} }}\int_0^\infty {e^{ - \alpha } \alpha ^n \,d\alpha } = \frac{{( - 1)^n n!}} {{(m + 1)^{n + 1} }}.$

--

As for the first integral, set $\displaystyle x=\frac1u.$

6. Originally Posted by Krizalid
$\displaystyle \int_0^1 {x^m \ln ^n x\,dx} = \frac{{( - 1)^n }} {{(m + 1)^{n + 1} }}\int_0^\infty {e^{ - \alpha } \alpha ^n \,d\alpha } = \frac{{( - 1)^n n!}} {{(m + 1)^{n + 1} }}.$
I see you are begining to love the gamma function

7. Aha, yes! It's really cool!

8. Originally Posted by Krizalid
Aha, yes! It's really cool!
I have a love hate relationship with the Gamma function.

9. Originally Posted by Mess
Hi people!

I can't solve these ones, they're tough!

Solve $\displaystyle \int {\frac{{\sqrt {\sqrt {x^4 + 1} - x^2 } }} {{x^4 + 1}}\,dx}$

Evaluate $\displaystyle \int_0^1 {x\ln x\,dx}$

Thanks!
And the answer (which may influence your decision to persist in working it out) is .............. :

10. I wouldn't trust in the integrator, after substitution $\displaystyle x=\frac1u$ leads another integral which can be solved easily.

11. Originally Posted by Krizalid
I wouldn't trust in the integrator, after substitution $\displaystyle x=\frac1u$ leads another integral which can be solved easily.
The Integrator can be trusted, just not trusted to give the answer in its simplest form.

-Dan

12. Originally Posted by topsquark
The Integrator can be trusted, just not trusted to give the answer in its simplest form.

-Dan
Mathematicians (the extremists) do not use calculator, computers, and machines.
Because they shall say this is a direct violation against what math is about.