What is the integral of (4sqrt(x))/(x^3+1) ?

The integral of (4sqrt(x))/(x^3+1) is 8/3 arctan(x^{3/2})+C

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

\int \frac{4 x}{x ^{3} + 1} d x

\frac{8}{3} arctan (x^{\frac{3}{2}}) + C

Solution steps

\int \frac{4 x}{x ^{3} + 1} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{x ^{3} + 1} d x

Apply u-substitution

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

= 4 \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

= 4 \cdot 2 \cdot \int \frac{u ^{2}}{u ^{6} + 1} d u

Apply u-substitution

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

= 4 \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

= 4 \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)

= 4 \cdot 2 \cdot \frac{1}{3} arctan (v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

= \frac{8}{3} arctan (x^{\frac{3}{2}}) + C

\int (4 x^{2} + 1)^{3} \cdot 8 x d x

x (1 - x^{2})

x \to 2 lim (x x^{2} - 4)

y^{^{''}} + 5 y^{^{'}} + y = 0

\int \frac{( 6 x ^{2} + 4 )}{( 3 y ^{2} - 4 )}

What is the integral of (4sqrt(x))/(x^3+1) ?
The integral of (4sqrt(x))/(x^3+1) is 8/3 arctan(x^{3/2})+C