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

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

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integral of (sqrt(x))/(4+x^3)

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

\frac{1}{3} arctan (\frac{1}{2} x^{\frac{3}{2}}) + C

Solution steps

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

Apply u-substitution

: \int \frac{2 u ^{2}}{4 + u ^{6}} d u

= \int \frac{2 u ^{2}}{4 + 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}}{4 + u ^{6}} d u

Apply u-substitution

: \int \frac{1}{3 ( 4 + v ^{2} )} d v

= 2 \cdot \int \frac{1}{3 ( 4 + 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}{4 + v ^{2}} d v

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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