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

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

\int \frac{3 x ^{3}}{1 - x ^{4}} d x

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

Solution steps

\int \frac{3 x ^{3}}{1 - x ^{4}} 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 ^{3}}{1 - x ^{4}} d x

Apply u-substitution

: \int - \frac{1}{4 u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 (- \frac{1}{4} ln ∣ u ∣)

Substitute back

u = 1 - x^{4}

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

Simplify

3 (- \frac{1}{4} ln 1 - x^{4}) : - \frac{3}{4} ln 1 - x^{4}

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

Add a constant to the solution

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

\int \frac{2 x ^{2} - x + 1}{x ^{2} - 5 x + 6} d x

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

\int \frac{5 - 2 x}{5 + 4 x - x ^{2}} d x

\int \frac{e ^{x}}{cos ^{2} ( e ^{x} )} d x

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

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