What is the integral of (x^5)/(1+x^3) ?

The integral of (x^5)/(1+x^3) is 1/3 (1+x^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 (x^5)/(1+x^3)

\int \frac{x ^{5}}{1 + x ^{3}} d x

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

Solution steps

\int \frac{x ^{5}}{1 + x ^{3}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{3} (u - ln ∣ u ∣)

Substitute back

u = 1 + x^{3}

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (\frac{y}{( x + y + z )})

\int_{- 2}^{3} \frac{31}{x ^{4}} d x

\frac{d}{d x} (10)

2 e^{3 x} \frac{d y}{d x} = - \frac{6 x}{y ^{2}}

4 x, 4, \frac{4}{x}, 0

What is the integral of (x^5)/(1+x^3) ?
The integral of (x^5)/(1+x^3) is 1/3 (1+x^3-ln|1+x^3|)+C