What is the integral of (3x^2)/((2x^3-4)^2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3x^2)/((2x^3-4)^2)

\int \frac{3 x ^{2}}{( 2 x ^{3} - 4 ) ^{2}} d x

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

Solution steps

\int \frac{3 x ^{2}}{( 2 x ^{3} - 4 ) ^{2}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x ^{2}}{( 2 x ^{3} - 4 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{6 u ^{2}} d u

= 3 \cdot \int \frac{1}{6 u ^{2}} d u

Take the constant out:

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

= 3 \cdot \frac{1}{6} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 3 \cdot \frac{1}{6} (- \frac{1}{u})

Substitute back

u = 2 x^{3} - 4

= 3 \cdot \frac{1}{6} (- \frac{1}{2 x ^{3} - 4})

Simplify

3 \cdot \frac{1}{6} (- \frac{1}{2 x ^{3} - 4}) : - \frac{1}{2 ( 2 x ^{3} - 4 )}

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

Add a constant to the solution

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

\frac{2 s - 1}{s ^{2} - 6 s + 18}

\frac{d}{d t} (12 t)

\int_{0}^{1} e^{- s t} d t

\int (x^{x}) d x

\frac{d y}{d x} = x^{5} (x - x y)

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