What is the integral of (7sqrt(x))/(x^3+1) ?

The integral of (7sqrt(x))/(x^3+1) is 14/3 arctan(x^{3/2})+C

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integral of (7sqrt(x))/(x^3+1)

\int \frac{7 x}{x ^{3} + 1} d x

\frac{14}{3} arctan (x^{\frac{3}{2}}) + C

Solution steps

\int \frac{7 x}{x ^{3} + 1} d x

Take the constant out:

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

= 7 \cdot \int \frac{x}{x ^{3} + 1} d x

Apply u-substitution

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

= 7 \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

= 7 \cdot 2 \cdot \int \frac{u ^{2}}{u ^{6} + 1} d u

Apply u-substitution

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

= 7 \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

= 7 \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)

= 7 \cdot 2 \cdot \frac{1}{3} arctan (v)

Substitute back

= 7 \cdot 2 \cdot \frac{1}{3} arctan ((x)^{3})

Simplify

7 \cdot 2 \cdot \frac{1}{3} arctan ((x)^{3}) : \frac{14}{3} arctan (x^{\frac{3}{2}})

= \frac{14}{3} arctan (x^{\frac{3}{2}})

Add a constant to the solution

= \frac{14}{3} arctan (x^{\frac{3}{2}}) + C

y = (x + 7) (7 x + 2)

\frac{d}{d x} (x^{3} (8 - 3 x))

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(s + 1) e^{- s π}

\frac{d y}{d x} = (x^{2})

What is the integral of (7sqrt(x))/(x^3+1) ?
The integral of (7sqrt(x))/(x^3+1) is 14/3 arctan(x^{3/2})+C