What is the integral of 6/(x^3)+16x^3 ?

The integral of 6/(x^3)+16x^3 is -3/(x^2)+4x^4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6/(x^3)+16x^3

\int \frac{6}{x ^{3}} + 16 x^{3} d x

- \frac{3}{x ^{2}} + 4 x^{4} + C

Solution steps

\int \frac{6}{x ^{3}} + 16 x^{3} 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{6}{x ^{3}} d x + \int 16 x^{3} d x

\int \frac{6}{x ^{3}} d x = - \frac{3}{x ^{2}}

\int 16 x^{3} d x = 4 x^{4}

= - \frac{3}{x ^{2}} + 4 x^{4}

Add a constant to the solution

= - \frac{3}{x ^{2}} + 4 x^{4} + C

\frac{\partial}{\partial t} (e^{t} cos (t))

\int tan^{3} (x) \cdot sec^{3} (x) d x

5 x^{3} 7 x, at x = 2

2 e^{- x} + e^{3 x}

\frac{\partial}{\partial x} (ln (x^{2} z^{2}))

What is the integral of 6/(x^3)+16x^3 ?
The integral of 6/(x^3)+16x^3 is -3/(x^2)+4x^4+C