What is the integral of (2/t-3/(t^2)) ?

The integral of (2/t-3/(t^2)) is 2ln|t|+3/t+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (2/t-3/(t^2))

\int (\frac{2}{t} - \frac{3}{t ^{2}}) d t

2 ln ∣ t ∣ + \frac{3}{t} + C

Solution steps

\int \frac{2}{t} - \frac{3}{t ^{2}} d t

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{2}{t} d t - \int \frac{3}{t ^{2}} d t

\int \frac{2}{t} d t = 2 ln ∣ t ∣

\int \frac{3}{t ^{2}} d t = - \frac{3}{t}

= 2 ln ∣ t ∣ - (- \frac{3}{t})

Simplify

= 2 ln ∣ t ∣ + \frac{3}{t}

Add a constant to the solution

= 2 ln ∣ t ∣ + \frac{3}{t} + C

What is the integral of (2/t-3/(t^2)) ?
The integral of (2/t-3/(t^2)) is 2ln|t|+3/t+C