What is the integral of 1/(sqrt(x)*(2+\sqrt[4]{x))} ?

The integral of 1/(sqrt(x)*(2+\sqrt[4]{x))} is 4(2+\sqrt[4]{x}-2ln|2+\sqrt[4]{x}|)+C

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integral of 1/(sqrt(x)*(2+\sqrt[4]{x))}

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

4 (2 + 4 x - 2 ln 2 + 4 x) + C

Solution steps

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

Apply u-substitution

= \int \frac{2}{2 + u} d u

Take the constant out:

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

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

Apply u-substitution

= 2 \cdot \int \frac{2 ( v - 2 )}{v} d v

Take the constant out:

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

= 2 \cdot 2 \cdot \int \frac{v - 2}{v} d v

Expand

\frac{v - 2}{v} : 1 - \frac{2}{v}

= 2 \cdot 2 \cdot \int 1 - \frac{2}{v} d v

Apply the Sum Rule:

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

= 2 \cdot 2 (\int 1 d v - \int \frac{2}{v} d v)

\int 1 d v = v

\int \frac{2}{v} d v = 2 ln ∣ v ∣

= 2 \cdot 2 (v - 2 ln ∣ v ∣)

Substitute back

= 2 \cdot 2 (2 + x - 2 ln 2 + x)

Simplify

2 \cdot 2 (2 + x - 2 ln 2 + x) : 4 (2 + 4 x - 2 ln 2 + 4 x)

= 4 (2 + 4 x - 2 ln 2 + 4 x)

Add a constant to the solution

= 4 (2 + 4 x - 2 ln 2 + 4 x) + C

\frac{\partial}{\partial y} (x + 4 y)

d er i v a t i v e \frac{( 2 x ^{2} tan ( x ) )}{sec ( x )}

x \to 0 lim (e^{x} cos (2 x))

\int_{2}^{e^{3}} \frac{1}{x} d x

d er i v a t i v e y = \frac{2 cos ( x )}{x + 1}

What is the integral of 1/(sqrt(x)*(2+\sqrt[4]{x))} ?
The integral of 1/(sqrt(x)*(2+\sqrt[4]{x))} is 4(2+\sqrt[4]{x}-2ln|2+\sqrt[4]{x}|)+C