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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

4 (4 x + 1 - ln 4 x + 1) + C

Solution steps

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

4 x^{3} = x^{\frac{3}{4}},

assuming

x \geq 0

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

Apply u-substitution

: \int \frac{4 u}{u + 1} d u

= \int \frac{4 u}{u + 1} d u

Take the constant out:

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

= 4 \cdot \int \frac{u}{u + 1} d u

Apply u-substitution

: \int \frac{v - 1}{v} d v

= 4 \cdot \int \frac{v - 1}{v} d v

Expand

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

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

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

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= 4 (v - ln ∣ v ∣)

Substitute back

= 4 (4 x + 1 - ln 4 x + 1)

Add a constant to the solution

= 4 (4 x + 1 - ln 4 x + 1) + C

\int \frac{cos ( 2 x )}{sin ^{4} ( 2 x )} d x

\int (6 x^{3} + \frac{1}{5 x ^{7}} - 4) d x

\int - 13 x^{- 37} d x

\int 1 + 3 y d y

\int \frac{6}{( x - 2 ) ^{2}} d x

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