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

The integral of x/((x+2)^{2/3)} is 3/4 (x+2)^{4/3}-6(x+2)^{1/3}+C

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

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

\frac{3}{4} (x + 2)^{\frac{4}{3}} - 6 (x + 2)^{\frac{1}{3}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 2}{u ^{\frac{2}{3}}} d u

= \int \frac{u - 2}{u ^{\frac{2}{3}}} d u

Expand

\frac{u - 2}{u ^{\frac{2}{3}}} : u^{\frac{1}{3}} - \frac{2}{u ^{\frac{2}{3}}}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{2}{u ^{\frac{2}{3}}} d u = 6 u^{\frac{1}{3}}

= \frac{3}{4} u^{\frac{4}{3}} - 6 u^{\frac{1}{3}}

Substitute back

u = x + 2

= \frac{3}{4} (x + 2)^{\frac{4}{3}} - 6 (x + 2)^{\frac{1}{3}}

Add a constant to the solution

= \frac{3}{4} (x + 2)^{\frac{4}{3}} - 6 (x + 2)^{\frac{1}{3}} + C

x \to 0 lim (cos (\frac{π}{2}))

y = 8 e^{x} cos (x)

\int \frac{( x + 1 )}{x} d x

\int \frac{1}{e ^{2 x} + 1} d x

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (x), y (0) = 1, y^{^{'}} (0) = - 1

What is the integral of x/((x+2)^{2/3)} ?
The integral of x/((x+2)^{2/3)} is 3/4 (x+2)^{4/3}-6(x+2)^{1/3}+C