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!
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.
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 $\displaystyle x=\frac1u.$