What is the integral of x^{-1}+sin(x)+e^x ?

The integral of x^{-1}+sin(x)+e^x is ln|x|-cos(x)+e^x+C

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integral of x^{-1}+sin(x)+e^x

\int x^{- 1} + sin (x) + e^{x} d x

ln ∣ x ∣ - cos (x) + e^{x} + C

Solution steps

\int x^{- 1} + sin (x) + e^{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 x^{- 1} d x + \int sin (x) d x + \int e^{x} d x

\int x^{- 1} d x = ln ∣ x ∣

\int sin (x) d x = - cos (x)

\int e^{x} d x = e^{x}

= ln ∣ x ∣ - cos (x) + e^{x}

Add a constant to the solution

= ln ∣ x ∣ - cos (x) + e^{x} + C

\frac{d y}{d x} + 2 x y = 1

h \to 0 - lim (\frac{- ln ( 1 + 6 h )}{h})

2 t^{3}

\int \frac{2}{cos ( 2 x )} d x

\frac{d}{d x} (\frac{x ^{2} + y - 1}{x + 1})

What is the integral of x^{-1}+sin(x)+e^x ?
The integral of x^{-1}+sin(x)+e^x is ln|x|-cos(x)+e^x+C