What is the integral of x^{11}sqrt(x^6+3) ?

The integral of x^{11}sqrt(x^6+3) is 1/15 (x^6-2)(x^6+3)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^{11}sqrt(x^6+3)

\int x^{11} x^{6} + 3 d x

\frac{1}{15} (x^{6} - 2) (x^{6} + 3)^{\frac{3}{2}} + C

Solution steps

\int x^{11} x^{6} + 3 d x

Apply u-substitution

: \int \frac{( u - 3 ) u}{6} d u

= \int \frac{( u - 3 ) u}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int (u - 3) u d u

Expand

(u - 3) u : u^{\frac{3}{2}} - 3 u

= \frac{1}{6} \cdot \int u^{\frac{3}{2}} - 3 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{1}{6} (\int u^{\frac{3}{2}} d u - \int 3 u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 3 u d u = 2 u^{\frac{3}{2}}

= \frac{1}{6} (\frac{2}{5} u^{\frac{5}{2}} - 2 u^{\frac{3}{2}})

Substitute back

u = x^{6} + 3

= \frac{1}{6} (\frac{2}{5} (x^{6} + 3)^{\frac{5}{2}} - 2 (x^{6} + 3)^{\frac{3}{2}})

Simplify

\frac{1}{6} (\frac{2}{5} (x^{6} + 3)^{\frac{5}{2}} - 2 (x^{6} + 3)^{\frac{3}{2}}) : \frac{1}{15} (x^{6} - 2) (x^{6} + 3)^{\frac{3}{2}}

= \frac{1}{15} (x^{6} - 2) (x^{6} + 3)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{15} (x^{6} - 2) (x^{6} + 3)^{\frac{3}{2}} + C

\frac{d y}{d x} = \frac{( x ^{2} - y ^{2} )}{x y}

\frac{d}{d x} (sech (e^{t a n (3 x)}))

\frac{\partial}{\partial x} (3 + cos (0.4 x - t))

8 x^{2}

\frac{d y}{d x} = \frac{y}{x} + e^{- \frac{y}{x}}

What is the integral of x^{11}sqrt(x^6+3) ?
The integral of x^{11}sqrt(x^6+3) is 1/15 (x^6-2)(x^6+3)^{3/2}+C