What is the integral of (9sqrt(x))/(x^3+1) ?

The integral of (9sqrt(x))/(x^3+1) is 6arctan(x^{3/2})+C

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integral of (9sqrt(x))/(x^3+1)

\int \frac{9 x}{x ^{3} + 1} d x

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

Solution steps

\int \frac{9 x}{x ^{3} + 1} d x

Take the constant out:

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

= 9 \cdot \int \frac{x}{x ^{3} + 1} d x

Apply u-substitution

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

= 9 \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

= 9 \cdot 2 \cdot \int \frac{u ^{2}}{u ^{6} + 1} d u

Apply u-substitution

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

= 9 \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

= 9 \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)

= 9 \cdot 2 \cdot \frac{1}{3} arctan (v)

Substitute back

= 9 \cdot 2 \cdot \frac{1}{3} arctan ((x)^{3})

Simplify

9 \cdot 2 \cdot \frac{1}{3} arctan ((x)^{3}) : 6 arctan (x^{\frac{3}{2}})

= 6 arctan (x^{\frac{3}{2}})

Add a constant to the solution

= 6 arctan (x^{\frac{3}{2}}) + C

\frac{\partial}{\partial x} (2 - (x^{2} + y^{2})^{\frac{1}{7}})

y = (x^{5} - 1)^{3}

x \to 0 lim (x sin (\frac{1}{∣ x ∣}))

- 2 e^{- x^{2}} x

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

What is the integral of (9sqrt(x))/(x^3+1) ?
The integral of (9sqrt(x))/(x^3+1) is 6arctan(x^{3/2})+C