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

The integral of (-1)/(t^2) is 1/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)/(t^2)

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

\frac{1}{t} + C

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{t}

= - (- \frac{1}{t})

Simplify

= \frac{1}{t}

Add a constant to the solution

= \frac{1}{t} + C

\int \frac{e ^{l n (x)}}{2 x ln ( x )} d x

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

\int \frac{t}{sin ^{2} ( t )} d t

\int \frac{cosh ( ln ( x ) )}{x} d x

\int x^{3} (x^{4} + 5)^{3} d x