อ้างอิง:
ข้อความเดิมเขียนโดยคุณ M@gpie
\[ \int \left( \ln (\ln x) + \frac{1}{(\ln x)^2}\right) dx \]
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Let $\displaystyle{x=e^u}$ then we get
$\displaystyle{\int \left( \ln (\ln x) + \frac{1}{(\ln x)^2}\right) dx=\int \left(e^u\ln u+\frac{e^u}{u^2}\right)du}$
Evaluate $\displaystyle{\int e^u\ln udu}$ with byparts method...
$\displaystyle{\int e^u\ln udu=e^u\ln u-\int\frac{e^u}{u}du+C=e^u\ln u-\frac{e^u}{u}-\int\frac{e^u}{u^2}du+C}$
$\therefore\displaystyle{\int \left(e^u\ln u+\frac{e^u}{u^2}\right)du=e^u\ln u-\frac{e^u}{u}+C}$
$\therefore\displaystyle{\int \left( \ln (\ln x) + \frac{1}{(\ln x)^2}\right) dx=x\left(\ln(\ln x)-\frac{1}{\ln x}\right)+C}$
__________________
$$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x-b\sin x}{a\sin x+b\cos x}dx=\ln\left(\frac{a}{b}\right)$$
BUT
$$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x+b\sin x}{a\sin x+b\cos x}dx=\frac{\pi ab}{a^{2}+b^{2}}+\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\ln\left(\frac{a}{b}\right)$$