What is the integral of [x^2+1/((3x)^2)] ?

The integral of [x^2+1/((3x)^2)] is (x^3)/3-1/(9x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of [x^2+1/((3x)^2)]

\int [x^{2} + \frac{1}{( 3 x ) ^{2}}] d x

\frac{x ^{3}}{3} - \frac{1}{9 x} + C

Solution steps

\int x^{2} + \frac{1}{( 3 x ) ^{2}} 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 x^{2} d x + \int \frac{1}{( 3 x ) ^{2}} d x

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

\int \frac{1}{( 3 x ) ^{2}} d x = - \frac{1}{9 x}

= \frac{x ^{3}}{3} - \frac{1}{9 x}

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{1}{9 x} + C

x \to 0 lim (x^{\frac{- 2}{3}})

f (x) = x^{3} - 3 x + 2, (2, 4)

\int sin (π - x) d x

\int - 5 e^{x} d x

\frac{d}{d x} (5 e^{2} + 2 π)

What is the integral of [x^2+1/((3x)^2)] ?
The integral of [x^2+1/((3x)^2)] is (x^3)/3-1/(9x)+C