What is the integral of x^5\sqrt[4]{1+x^3} ?

The integral of x^5\sqrt[4]{1+x^3} is 4/3 (1/9 (1+x^3)^{9/4}-1/5 (1+x^3)^{5/4})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^5\sqrt[4]{1+x^3}

\int x^{5} 4 1 + x^{3} d x

\frac{4}{3} (\frac{1}{9} (1 + x^{3})^{\frac{9}{4}} - \frac{1}{5} (1 + x^{3})^{\frac{5}{4}}) + C

Solution steps

\int x^{5} 4 1 + x^{3} d x

Apply u-substitution

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

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

Expand

u^{4} (u^{4} - 1) : u^{8} - u^{4}

= \frac{4}{3} \cdot \int u^{8} - u^{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

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

\int u^{8} d u = \frac{u ^{9}}{9}

\int u^{4} d u = \frac{u ^{5}}{5}

= \frac{4}{3} (\frac{u ^{9}}{9} - \frac{u ^{5}}{5})

Substitute back

u = 4 1 + x^{3}

= \frac{4}{3} (\frac{( 4 1 + x ^{3} ) ^{9}}{9} - \frac{( 4 1 + x ^{3} ) ^{5}}{5})

Simplify

\frac{4}{3} (\frac{( 4 1 + x ^{3} ) ^{9}}{9} - \frac{( 4 1 + x ^{3} ) ^{5}}{5}) : \frac{4}{3} (\frac{1}{9} (1 + x^{3})^{\frac{9}{4}} - \frac{1}{5} (1 + x^{3})^{\frac{5}{4}})

= \frac{4}{3} (\frac{1}{9} (1 + x^{3})^{\frac{9}{4}} - \frac{1}{5} (1 + x^{3})^{\frac{5}{4}})

Add a constant to the solution

= \frac{4}{3} (\frac{1}{9} (1 + x^{3})^{\frac{9}{4}} - \frac{1}{5} (1 + x^{3})^{\frac{5}{4}}) + C

\int 2 x ln (3) d x

\int \frac{7 x ^{2}}{1 - x ^{3}} d x

\int (2 cos (4 x) - 1)^{\frac{1}{2}} sin (4 x) d x

\int \frac{1}{1 - x ^{2}} \cdot x d x

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

What is the integral of x^5\sqrt[4]{1+x^3} ?
The integral of x^5\sqrt[4]{1+x^3} is 4/3 (1/9 (1+x^3)^{9/4}-1/5 (1+x^3)^{5/4})+C