What is the integral of (1/(3x^3)) ?

The integral of (1/(3x^3)) is -1/(6x^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1/(3x^3))

\int (\frac{1}{3 x ^{3}}) d x

- \frac{1}{6 x ^{2}} + C

Solution steps

\int \frac{1}{3 x ^{3}} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{x ^{3}} d x

Apply the Power Rule

: - \frac{1}{2 x ^{2}}

= \frac{1}{3} (- \frac{1}{2 x ^{2}})

Simplify

\frac{1}{3} (- \frac{1}{2 x ^{2}}) : - \frac{1}{6 x ^{2}}

= - \frac{1}{6 x ^{2}}

Add a constant to the solution

= - \frac{1}{6 x ^{2}} + C

\int \frac{1}{( arctan ( \frac{y}{6} ) ) ( 36 + y ^{2} )} d y

\int \frac{x + 4}{x ^{2} - 5 x + 6} d x

\int \frac{x ^{7} e ^{x} - 2 x ^{6}}{x ^{7}} d x

\int 4 (x^{2} + 3 x + 7) (2 x + 3) d x

\int \frac{csc ^{2} ( y )}{cot ^{5} ( y )} d y