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{10}{x ^{3} + x ^{2}} d x

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

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

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

\int x^{2} + 8 x^{5} d x

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