What is the integral of (8sqrt(x))/(x^3+1) ?

The integral of (8sqrt(x))/(x^3+1) is 16/3 arctan(x^{3/2})+C

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integral of (8sqrt(x))/(x^3+1)

\int \frac{8 x}{x ^{3} + 1} d x

\frac{16}{3} arctan (x^{\frac{3}{2}}) + C

Solution steps

\int \frac{8 x}{x ^{3} + 1} d x

Take the constant out:

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

= 8 \cdot \int \frac{x}{x ^{3} + 1} d x

Apply u-substitution

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

= 8 \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

= 8 \cdot 2 \cdot \int \frac{u ^{2}}{u ^{6} + 1} d u

Apply u-substitution

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

= 8 \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

= 8 \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)

= 8 \cdot 2 \cdot \frac{1}{3} arctan (v)

Substitute back

= 8 \cdot 2 \cdot \frac{1}{3} arctan ((x)^{3})

Simplify

8 \cdot 2 \cdot \frac{1}{3} arctan ((x)^{3}) : \frac{16}{3} arctan (x^{\frac{3}{2}})

= \frac{16}{3} arctan (x^{\frac{3}{2}})

Add a constant to the solution

= \frac{16}{3} arctan (x^{\frac{3}{2}}) + C

x sin (π x)

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\frac{d y}{d x} = x x - 30

\frac{d}{d x} (2 sin (x) \cdot cos (x))

0

What is the integral of (8sqrt(x))/(x^3+1) ?
The integral of (8sqrt(x))/(x^3+1) is 16/3 arctan(x^{3/2})+C