What is the integral of 5e^{2x}+1/x ?

The integral of 5e^{2x}+1/x is 5/2 e^{2x}+ln|x|+C

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integral of 5e^{2x}+1/x

\int 5 e^{2 x} + \frac{1}{x} d x

\frac{5}{2} e^{2 x} + ln ∣ x ∣ + C

Solution steps

\int 5 e^{2 x} + \frac{1}{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 5 e^{2 x} d x + \int \frac{1}{x} d x

\int 5 e^{2 x} d x = \frac{5}{2} e^{2 x}

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{5}{2} e^{2 x} + ln ∣ x ∣

Add a constant to the solution

= \frac{5}{2} e^{2 x} + ln ∣ x ∣ + C

\frac{d}{d x} (\frac{1 + csc ( x ^{4} )}{1 - cot ( x ^{4} )})

\frac{\partial}{\partial x} (\frac{e ^{x y}}{x + 1})

\int u e^{- u} d u

y = 4 ln (x), at x = 4

\int_{0}^{1} (x^{\frac{4}{5}}) d x

What is the integral of 5e^{2x}+1/x ?
The integral of 5e^{2x}+1/x is 5/2 e^{2x}+ln|x|+C