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

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

\int (\frac{x ^{3}}{x ^{4} + 9}) d x

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

Solution steps

\int \frac{x ^{3}}{x ^{4} + 9} 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 = x^{4} + 9

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

Add a constant to the solution

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

\int [x (x^{2} - 16)^{2} - \frac{1}{2 x + 5}] d x

\int \frac{3 x + 5}{x ^{2} + 4} d x

\int t^{3} t^{4} + 1 d t

\int \frac{cos ( x )}{sin ^{3} ( x )} d x

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