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!

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- Dec 17th 2007, 07:09 AMMessIntegrals!
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! - Dec 17th 2007, 07:19 AMbobak
for the second one do integration by parts, you should get an answer of

$\displaystyle \frac{x^2 lnx}{2} - \frac{x^2}{4}$ - Dec 17th 2007, 07:44 AMTKHunny
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.

This leads me to ask two things:

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. - Dec 17th 2007, 08:20 AMcolby2152
- Dec 17th 2007, 08:48 AMKrizalid
Here's a specific calculation in this case.

$\displaystyle \ln x=\int_1^x\frac1u\,du.$ 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 $\displaystyle x=\frac1u.$ - Dec 17th 2007, 09:33 AMThePerfectHacker
- Dec 17th 2007, 10:12 AMKrizalid
Aha, yes! It's really cool! :D

- Dec 17th 2007, 10:24 AMcolby2152
- Dec 17th 2007, 02:08 PMmr fantastic
- Dec 17th 2007, 02:43 PMKrizalid
I wouldn't trust in the integrator, after substitution $\displaystyle x=\frac1u$ leads another integral which can be solved easily.

- Dec 18th 2007, 07:16 PMtopsquark
- Dec 18th 2007, 07:17 PMThePerfectHacker