What is the integral of 4x^2+1/((2x)^2) ?

The integral of 4x^2+1/((2x)^2) is (4x^3)/3+1/(4x)-1/(2x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4x^2+1/((2x)^2)

\int 4 x^{2} + \frac{1}{( 2 x ) ^{2}} d x

\frac{4 x ^{3}}{3} + \frac{1}{4 x} - \frac{1}{2 x} + C

Solution steps

\int 4 x^{2} + \frac{1}{( 2 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 4 x^{2} d x + \int \frac{1}{( 2 x ) ^{2}} d x

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int \frac{1}{( 2 x ) ^{2}} d x = \frac{1}{4 x} - \frac{1}{2 x}

= \frac{4 x ^{3}}{3} + \frac{1}{4 x} - \frac{1}{2 x}

Add a constant to the solution

= \frac{4 x ^{3}}{3} + \frac{1}{4 x} - \frac{1}{2 x} + C

2 t^{4}

x \frac{d y}{d x} = y + 2 x e^{- \frac{y}{x}}

\frac{6 s}{( s ^{2} - 8 s + 25 )}

\int \frac{6}{x ln ( 7 x )} d x

\frac{d y}{d x} = x - 2.5 y

What is the integral of 4x^2+1/((2x)^2) ?
The integral of 4x^2+1/((2x)^2) is (4x^3)/3+1/(4x)-1/(2x)+C