What is the integral of (2sqrt(x))/(x^3+1) ?

The integral of (2sqrt(x))/(x^3+1) is 4/3 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 (2sqrt(x))/(x^3+1)

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

\frac{4}{3} arctan (x^{\frac{3}{2}}) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

2 \cdot 2 \cdot \frac{1}{3} arctan ((x)^{3}) : \frac{4}{3} arctan (x^{\frac{3}{2}})

= \frac{4}{3} arctan (x^{\frac{3}{2}})

Add a constant to the solution

= \frac{4}{3} arctan (x^{\frac{3}{2}}) + C

\int e^{\frac{t}{10}} d t

\int \frac{tan ( x )}{cos ( x )} d x

14 y^{^{''}} + y^{^{'}} + y = x^{2} - 2 x

\frac{\partial}{\partial y} (2 x y^{2})

\int_{0}^{8 π} x^{2} sin (2 x) d x

What is the integral of (2sqrt(x))/(x^3+1) ?
The integral of (2sqrt(x))/(x^3+1) is 4/3 arctan(x^{3/2})+C