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

The integral of 4 x/((x+2)^{(1/4))} is 4(4/7 (x+2)^{7/4}-8/3 (x+2)^{3/4})+C

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integral of 4 x/((x+2)^{(1/4))}

\int 4 \frac{x}{( x + 2 ) ^{(\frac{1}{4})}} d x

4 (\frac{4}{7} (x + 2)^{\frac{7}{4}} - \frac{8}{3} (x + 2)^{\frac{3}{4}}) + C

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{u - 2}{u ^{\frac{1}{4}}} : u^{\frac{3}{4}} - \frac{2}{u ^{\frac{1}{4}}}

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

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

\int u^{\frac{3}{4}} d u = \frac{4}{7} u^{\frac{7}{4}}

\int \frac{2}{u ^{\frac{1}{4}}} d u = \frac{8}{3} u^{\frac{3}{4}}

= 4 (\frac{4}{7} u^{\frac{7}{4}} - \frac{8}{3} u^{\frac{3}{4}})

Substitute back

u = x + 2

= 4 (\frac{4}{7} (x + 2)^{\frac{7}{4}} - \frac{8}{3} (x + 2)^{\frac{3}{4}})

Add a constant to the solution

= 4 (\frac{4}{7} (x + 2)^{\frac{7}{4}} - \frac{8}{3} (x + 2)^{\frac{3}{4}}) + C

x \to 1 - lim (3^{\frac{1}{x}})

\frac{\partial}{\partial y} (e^{x} + cos (y + z))

(x^{2} + 4) y^{^{'}} + 3 x y = x

n = 1 \sum \infty \frac{7}{4 ^{n}}

x \to 0 lim (\frac{e ^{a x} - 1}{x})

What is the integral of 4 x/((x+2)^{(1/4))} ?
The integral of 4 x/((x+2)^{(1/4))} is 4(4/7 (x+2)^{7/4}-8/3 (x+2)^{3/4})+C