What is the integral of \sqrt[4]{(1+x^{1/2})^3} ?

The integral of \sqrt[4]{(1+x^{1/2})^3} is 8/77 (7sqrt(x)-4)(sqrt(x)+1)^{7/4}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of \sqrt[4]{(1+x^{1/2})^3}

\frac{8}{77} (7 x - 4) (x + 1)^{\frac{7}{4}} + C

Solution steps

Apply u-substitution

Take the constant out:

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

Expand

= \frac{2}{3} \cdot \int \frac{1}{u ^{\frac{1}{12}}} - \frac{1}{u ^{\frac{5}{12}}} 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{2}{3} (\int \frac{1}{u ^{\frac{1}{12}}} d u - \int \frac{1}{u ^{\frac{5}{12}}} d u)

\int \frac{1}{u ^{\frac{1}{12}}} d u = \frac{12}{11} u^{\frac{11}{12}}

\int \frac{1}{u ^{\frac{5}{12}}} d u = \frac{12}{7} u^{\frac{7}{12}}

= \frac{2}{3} (\frac{12}{11} u^{\frac{11}{12}} - \frac{12}{7} u^{\frac{7}{12}})

Substitute back

u = (1 + x^{\frac{1}{2}})^{3}

= \frac{2}{3} (\frac{12}{11} ((1 + x^{\frac{1}{2}})^{3})^{\frac{11}{12}} - \frac{12}{7} ((1 + x^{\frac{1}{2}})^{3})^{\frac{7}{12}})

Simplify

\frac{2}{3} (\frac{12}{11} ((1 + x^{\frac{1}{2}})^{3})^{\frac{11}{12}} - \frac{12}{7} ((1 + x^{\frac{1}{2}})^{3})^{\frac{7}{12}}) : \frac{8}{77} (7 x - 4) (x + 1)^{\frac{7}{4}}

= \frac{8}{77} (7 x - 4) (x + 1)^{\frac{7}{4}}

Add a constant to the solution

= \frac{8}{77} (7 x - 4) (x + 1)^{\frac{7}{4}} + C

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

\int \frac{5}{4 x ^{2} + 1} d x

\int (3 - y) cos (x)

\int \frac{x + 4}{2 x ^{3} - 18 x} d x

\int x sin (6 x^{2}) d x

What is the integral of \sqrt[4]{(1+x^{1/2})^3} ?
The integral of \sqrt[4]{(1+x^{1/2})^3} is 8/77 (7sqrt(x)-4)(sqrt(x)+1)^{7/4}+C