What is the integral of (x^3)/(9-x^4) ?

The integral of (x^3)/(9-x^4) is -1/4 ln|9-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 (x^3)/(9-x^4)

\int \frac{x ^{3}}{9 - x ^{4}} d x

- \frac{1}{4} ln 9 - x^{4} + C

Solution steps

\int \frac{x ^{3}}{9 - x ^{4}} d x

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{4} ln ∣ u ∣

Substitute back

u = 9 - x^{4}

= - \frac{1}{4} ln 9 - x^{4}

Add a constant to the solution

= - \frac{1}{4} ln 9 - x^{4} + C

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

t \to 0 + lim (\frac{sin ( t )}{t})

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

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

n = 0 \sum \infty \frac{e}{n}

What is the integral of (x^3)/(9-x^4) ?
The integral of (x^3)/(9-x^4) is -1/4 ln|9-x^4|+C