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

The integral of (4x^2)/(1+x^3) is 4/3 ln|1+x^3|+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+x^3)

\int \frac{4 x ^{2}}{1 + x ^{3}} d x

\frac{4}{3} ln 1 + x^{3} + C

Solution steps

\int \frac{4 x ^{2}}{1 + x ^{3}} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{2}}{1 + x ^{3}} d x

Apply u-substitution

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

= 4 \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

= 4 \cdot \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{3} ln ∣ u ∣

Substitute back

u = 1 + x^{3}

= 4 \cdot \frac{1}{3} ln 1 + x^{3}

Simplify

4 \cdot \frac{1}{3} ln 1 + x^{3} : \frac{4}{3} ln 1 + x^{3}

= \frac{4}{3} ln 1 + x^{3}

Add a constant to the solution

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

\int 3 y 3 2 y^{2} - 8 d y

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

f (x) = 0.3 x^{1.2}

a - b y^{2} = \frac{d y}{d x}

(3.9) (- 5.3)

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