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

The integral of (3x^2)/((1-x)(x^2+x+1)) is -ln(|1-x||x^2+x+1|)+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)/((1-x)(x^2+x+1))

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

- ln (∣ 1 - x ∣ x^{2} + x + 1) + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{3 u} d u

= 3 \cdot \int - \frac{1}{3 u} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 (- \frac{1}{3} ln ∣ u ∣)

Substitute back

u = (1 - x) (x^{2} + x + 1)

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

Simplify

3 (- \frac{1}{3} ln (1 - x) (x^{2} + x + 1)) : - ln (∣ 1 - x ∣ x^{2} + x + 1)

= - ln (∣ 1 - x ∣ x^{2} + x + 1)

Add a constant to the solution

= - ln (∣ 1 - x ∣ x^{2} + x + 1) + C

x \to 4 lim (- x^{2} + 6 x - 3)

n = 0 \sum \infty x^{2 n + 1}

\frac{d}{d x} (\frac{1}{1 - 2 x})

y = x sin (x) + 2

x^{^{''}} - x^{^{'}} - 6 x = 6 e^{3 t} + 2 e^{- 2 t}

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