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

The integral of (sqrt(x))/(1+x^3) is 2/3 arctan(x^{3/2})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sqrt(x))/(1+x^3)

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

\frac{2}{3} arctan (x^{\frac{3}{2}}) + C

Solution steps

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

Apply u-substitution

: \int \frac{2 u ^{2}}{1 + u ^{6}} d u

= \int \frac{2 u ^{2}}{1 + u ^{6}} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2}}{1 + u ^{6}} d u

Apply u-substitution

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

= 2 \cdot \int \frac{1}{3 ( 1 + v ^{2} )} d v

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int \frac{1}{1 + v ^{2}} d v

Use the common integral:

\int \frac{1}{1 + v ^{2}} d v = arctan (v)

= 2 \cdot \frac{1}{3} arctan (v)

Substitute back

= 2 \cdot \frac{1}{3} arctan ((x)^{3})

Simplify

2 \cdot \frac{1}{3} arctan ((x)^{3}) : \frac{2}{3} arctan (x^{\frac{3}{2}})

= \frac{2}{3} arctan (x^{\frac{3}{2}})

Add a constant to the solution

= \frac{2}{3} arctan (x^{\frac{3}{2}}) + C

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - y = 20 x e^{x}

\frac{d y}{d x} + y = x

f (x) = 2 x^{2} - 5, (2, - 2)

\int \frac{1}{x x ^{4} - 16} d x

\frac{d}{d x} (ln (x) + ln (sin (y)))

What is the integral of (sqrt(x))/(1+x^3) ?
The integral of (sqrt(x))/(1+x^3) is 2/3 arctan(x^{3/2})+C