What is the integral of x^{11}sqrt(x^6+4) ?

The integral of x^{11}sqrt(x^6+4) is 1/45 (3x^6-8)(x^6+4)^{3/2}+C

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integral of x^{11}sqrt(x^6+4)

\int x^{11} x^{6} + 4 d x

\frac{1}{45} (3 x^{6} - 8) (x^{6} + 4)^{\frac{3}{2}} + C

Solution steps

\int x^{11} x^{6} + 4 d x

Apply u-substitution

: \int \frac{( u - 4 ) u}{6} d u

= \int \frac{( u - 4 ) 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 - 4) u d u

Expand

(u - 4) u : u^{\frac{3}{2}} - 4 u

= \frac{1}{6} \cdot \int u^{\frac{3}{2}} - 4 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 4 u d u)

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

\int 4 u d u = \frac{8}{3} u^{\frac{3}{2}}

= \frac{1}{6} (\frac{2}{5} u^{\frac{5}{2}} - \frac{8}{3} u^{\frac{3}{2}})

Substitute back

u = x^{6} + 4

= \frac{1}{6} (\frac{2}{5} (x^{6} + 4)^{\frac{5}{2}} - \frac{8}{3} (x^{6} + 4)^{\frac{3}{2}})

Simplify

\frac{1}{6} (\frac{2}{5} (x^{6} + 4)^{\frac{5}{2}} - \frac{8}{3} (x^{6} + 4)^{\frac{3}{2}}) : \frac{1}{45} (3 x^{6} - 8) (x^{6} + 4)^{\frac{3}{2}}

= \frac{1}{45} (3 x^{6} - 8) (x^{6} + 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{45} (3 x^{6} - 8) (x^{6} + 4)^{\frac{3}{2}} + C

\int 7 v (v^{2} + 2)^{2} d v

\frac{d}{d θ} (3 + 2 sin (θ))

f (x) = \frac{x + x}{x ^{2}}

- \frac{14}{s ^{5}}

f (x) = 3 4 - 9 x

What is the integral of x^{11}sqrt(x^6+4) ?
The integral of x^{11}sqrt(x^6+4) is 1/45 (3x^6-8)(x^6+4)^{3/2}+C