What is the integral of x^7sqrt(x^4+6) ?

The integral of x^7sqrt(x^4+6) is 1/10 (x^4-4)(x^4+6)^{3/2}+C

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integral of x^7sqrt(x^4+6)

\int x^{7} x^{4} + 6 d x

\frac{1}{10} (x^{4} - 4) (x^{4} + 6)^{\frac{3}{2}} + C

Solution steps

\int x^{7} x^{4} + 6 d x

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

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

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

\int 6 u d u = 4 u^{\frac{3}{2}}

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

Substitute back

u = x^{4} + 6

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

Simplify

\frac{1}{4} (\frac{2}{5} (x^{4} + 6)^{\frac{5}{2}} - 4 (x^{4} + 6)^{\frac{3}{2}}) : \frac{1}{10} (x^{4} - 4) (x^{4} + 6)^{\frac{3}{2}}

= \frac{1}{10} (x^{4} - 4) (x^{4} + 6)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{10} (x^{4} - 4) (x^{4} + 6)^{\frac{3}{2}} + C

(x^{3} - sec (3 x^{2} - 4))^{- 3}

\frac{d}{d x} (\frac{2}{x} - 1)

y^{^{''}} + 3 y^{^{'}} + 2 y = 0, y (1) = 1, y^{^{'}} (1) = 0

- \frac{8}{x ^{2}} - \frac{8}{x ^{3}}

\int \frac{x ^{2} + x}{x ^{4} - 3 x ^{2} - 4} d x

What is the integral of x^7sqrt(x^4+6) ?
The integral of x^7sqrt(x^4+6) is 1/10 (x^4-4)(x^4+6)^{3/2}+C