What is the integral of-1/(2t) ?

The integral of-1/(2t) is -1/2 ln|t|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of-1/(2t)

\int - \frac{1}{2 t} d t

- \frac{1}{2} ln ∣ t ∣ + C

Solution steps

\int - \frac{1}{2 t} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot \int \frac{1}{t} d t

Use the common integral:

\int \frac{1}{t} d t = ln (∣ t ∣)

= - \frac{1}{2} ln ∣ t ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ t ∣ + C

\int 6 (1 + tan^{2} (x)) d x

\int (- 2)^{x} d x

\int (1 - x)^{10} d x

\int \frac{x + 2}{( x - 4 ) ^{2}} d x

\int \frac{2}{x} + 1 d x