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

The integral of x^{11}sqrt(x^6+6) is 1/3 (1/5 (x^6+6)^{5/2}-2(x^6+6)^{3/2})+C

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{( u ^{2} - 6 ) u ^{2}}{3} d u

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

Take the constant out:

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

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

Expand

(u^{2} - 6) u^{2} : u^{4} - 6 u^{2}

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

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

\int 6 u^{2} d u = 2 u^{3}

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

Substitute back

u = x^{6} + 6

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

Simplify

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

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

Add a constant to the solution

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

2 x^{2} - 3 x + 4

x \to 1 lim (\frac{x - 2}{x - 1})

x \to 0 + lim (ln (x) tan (x))

x \to - 4 lim (\frac{2}{( x + 4 ) ^{6}})

\frac{d}{d x} f (x) = (x^{4} + 3 x^{2} - 9)^{6}

What is the integral of x^{11}sqrt(x^6+6) ?
The integral of x^{11}sqrt(x^6+6) is 1/3 (1/5 (x^6+6)^{5/2}-2(x^6+6)^{3/2})+C