What is the integral of x/((x^2+2)^{4/3)} ?

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

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integral of x/((x^2+2)^{4/3)}

\int \frac{x}{( x ^{2} + 2 ) ^{\frac{4}{3}}} d x

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

Solution steps

\int \frac{x}{( x ^{2} + 2 ) ^{\frac{4}{3}}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{\frac{4}{3}}} d u

= \int \frac{1}{2 u ^{\frac{4}{3}}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{\frac{4}{3}}} d u

Apply the Power Rule

: - \frac{3}{u ^{\frac{1}{3}}}

= \frac{1}{2} (- \frac{3}{u ^{\frac{1}{3}}})

Substitute back

u = x^{2} + 2

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

Simplify

\frac{1}{2} (- \frac{3}{( x ^{2} + 2 ) ^{\frac{1}{3}}}) : - \frac{3}{2 ( x ^{2} + 2 ) ^{\frac{1}{3}}}

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

Add a constant to the solution

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

y^{^{'}} = 3 y - y^{2}

c x

y^{^{'}} - \frac{4}{x} y = \frac{y ^{3}}{x ^{3}}, y (1) = 1

\frac{d}{d x} (\frac{7}{10} x^{2} + 4 x^{3})

\frac{d y}{d x} - y = - \frac{x}{2}

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