What is the integral of 1/x+zcos(xz) ?

The integral of 1/x+zcos(xz) is ln|x|+sin(zx)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/x+zcos(xz)

\int \frac{1}{x} + z cos (x z) d x

ln ∣ x ∣ + sin (z x) + C

Solution steps

\int \frac{1}{x} + z cos (x z) 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 cos (z x) z d x

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ + sin (z x)

Add a constant to the solution

= ln ∣ x ∣ + sin (z x) + C

\int (\frac{tan ( x )}{cot ( x )} + 1) d x

\int \frac{4 x sin ( x ) - 5 csc ( x )}{sin ( x )} d x

\int 6 x^{3} e^{x^{2} + 3} d x

\int \frac{x ^{2} + 49}{3 x ^{2}} d x

\int 5 x arcsin (x) d x

What is the integral of 1/x+zcos(xz) ?
The integral of 1/x+zcos(xz) is ln|x|+sin(zx)+C