Exhausting integration of a physical function I am to find the area under a curve in particular interval. The equation of the curve is:
$$u(\lambda) = \frac{2c^2h}{\lambda^5(e^{\frac{hc}{\lambda kT}}-1)}$$
The problem is, I am absolutely stuck. I can't even imagine integrating such function. Is there any other way to bite this?
 A: HINT:
$$\int_{a}^{b}\frac{2\text{c}^2\text{h}}{\lambda^5\left(\exp\left[\frac{\text{ch}}{\lambda\text{KT}}\right]-1\right)}\space\text{d}\lambda=2\text{c}^2\text{h}\int_{a}^{b}\frac{1}{\lambda^5\left(\exp\left[\frac{\text{ch}}{\lambda\text{KT}}\right]-1\right)}\space\text{d}\lambda=$$

Subsitute $u=\exp\left[\frac{\text{ch}}{\lambda\text{KT}}\right]$ and $\text{d}u=-\frac{\text{ch}\exp\left[\frac{\text{ch}}{\lambda\text{KT}}\right]}{\text{kT}\lambda^2}\space\text{d}\lambda$:
This gives a new lower bound $u=\exp\left[\frac{\text{ch}}{a\text{KT}}\right]$ and upper bound $u=\exp\left[\frac{\text{ch}}{b\text{KT}}\right]$:

$$-\frac{2\text{K}^4\text{T}^4}{\text{c}^2\text{h}^3}\int_{\exp\left[\frac{\text{ch}}{a\text{KT}}\right]}^{\exp\left[\frac{\text{ch}}{b\text{KT}}\right]}\frac{\ln^3(u)}{u(u-1)}\space\text{d}u=$$

Subsitute $s=\ln(u)$ and $\text{d}s=\frac{1}{u}\space\text{d}u$:
This gives a new lower bound $s=\frac{\text{ch}}{a\text{KT}}$ and upper bound $s=\frac{\text{ch}}{b\text{KT}}$:

$$-\frac{2\text{K}^4\text{T}^4}{\text{c}^2\text{h}^3}\int_{\frac{\text{ch}}{a\text{KT}}}^{\frac{\text{ch}}{b\text{KT}}}\frac{s^3}{\exp(s)-1}\space\text{d}s$$
