What is the integral of x^{11}sqrt(x^6+2) ?

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

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integral of x^{11}sqrt(x^6+2)

\int x^{11} x^{6} + 2 d x

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

Solution steps

\int x^{11} x^{6} + 2 d x

Apply u-substitution

: \int \frac{( u - 2 ) u}{6} d u

= \int \frac{( u - 2 ) 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 - 2) u d u

Expand

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

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

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

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

= \frac{1}{6} (\frac{2}{5} u^{\frac{5}{2}} - \frac{4}{3} u^{\frac{3}{2}})

Substitute back

u = x^{6} + 2

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

Simplify

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

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

Add a constant to the solution

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

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