What is the integral of (sqrt(x))/(1+\sqrt[4]{x^3)} ?

The integral of (sqrt(x))/(1+\sqrt[4]{x^3)} is 4/3 (1+x^{3/4}-ln|1+x^{3/4}|)+C

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integral of (sqrt(x))/(1+\sqrt[4]{x^3)}

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

Solution steps

assuming

x \geq 0

= \int \frac{x}{1 + x ^{\frac{3}{4}}} d x

Apply u-substitution

: \int \frac{4 ( u - 1 )}{3 u} d u

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

Take the constant out:

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

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

Expand

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

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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{4}{3} (u - ln ∣ u ∣)

Substitute back

u = 1 + x^{\frac{3}{4}}

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

Add a constant to the solution

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

\int_{2}^{\infty} e^{- 4 x} d x

\frac{\partial}{\partial x} (x y e^{x^{2} + y^{2}})

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

x \to 1 lim (\frac{x}{x ^{2} + 2})

f (x) = \frac{3 π x ^{2}}{2}

What is the integral of (sqrt(x))/(1+\sqrt[4]{x^3)} ?
The integral of (sqrt(x))/(1+\sqrt[4]{x^3)} is 4/3 (1+x^{3/4}-ln|1+x^{3/4}|)+C