- \(\displaystyle \int \frac {xe^x\ln x - e^x}{x\ln^2 x} = \color {blue} {\frac {e^x}{\ln x}}\)
- Used 2018
- \(\displaystyle \int (\tan(x) + \cot(x))^2 ~dx = \color {blue} {\tan(x)-\cot(x)}\)
Solution: \(\sin ^2 (x) + \cos ^2 (x) = 1\)
Used 2018
Solution
1. \(\int \frac{x e^x \ln x - e^x}{x (\ln x)^2} \, dx = \frac{e^x}{\ln x} + C\).
1.1. The integrand is the derivative of the quotient \(g(x) = \frac{e^x}{\ln x}\).
1.2. By the quotient rule: \[\frac{d}{dx}\left(\frac{e^x}{\ln x}\right) = \frac{\left(\frac{d}{dx} e^x\right) \ln x - e^x \left(\frac{d}{dx} \ln x\right)}{(\ln x)^2} = \frac{e^x \ln x - e^x \cdot \frac{1}{x}}{(\ln x)^2} = \frac{x e^x \ln x - e^x}{x (\ln x)^2}.\]
1.3. Since the integrand is the exact derivative of \(\frac{e^x}{\ln x}\), the antiderivative is \(\frac{e^x}{\ln x} + C\).
2. \(\int (\tan x + \cot x)^2 \, dx = \tan(x) - \cot(x) + C\).
2.1. Expand the square: \[(\tan x + \cot x)^2 = \tan^2 x + 2\tan x \cot x + \cot^2 x = \tan^2 x + 2 + \cot^2 x.\]
2.2. Apply the Pythagorean trigonometric identities \(\tan^2 x = \sec^2 x - 1\) and \(\cot^2 x = \csc^2 x - 1\): \[\tan^2 x + 2 + \cot^2 x = (\sec^2 x - 1) + 2 + (\csc^2 x - 1) = \sec^2 x + \csc^2 x.\]
2.3. Integrate term-by-term: \[\int (\sec^2 x + \csc^2 x) \, dx = \tan x - \cot x + C.\] Q.E.D.