What is the integral of (1/x+4sin(2x)+e^x) ?

The integral of (1/x+4sin(2x)+e^x) is ln|x|-2cos(2x)+e^x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (1/x+4sin(2x)+e^x)

\int (\frac{1}{x} + 4 sin (2 x) + e^{x}) d x

ln ∣ x ∣ - 2 cos (2 x) + e^{x} + C

Solution steps

\int \frac{1}{x} + 4 sin (2 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 \frac{1}{x} d x + \int 4 sin (2 x) d x + \int e^{x} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int 4 sin (2 x) d x = - 2 cos (2 x)

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

= ln ∣ x ∣ - 2 cos (2 x) + e^{x}

Add a constant to the solution

= ln ∣ x ∣ - 2 cos (2 x) + e^{x} + C

What is the integral of (1/x+4sin(2x)+e^x) ?
The integral of (1/x+4sin(2x)+e^x) is ln|x|-2cos(2x)+e^x+C