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

The integral of x^7sqrt(x^4+7) is 1/30 (3x^4-14)(x^4+7)^{3/2}+C

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

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

\frac{1}{30} (3 x^{4} - 14) (x^{4} + 7)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{( u - 7 ) u}{4} d u

= \int \frac{( u - 7 ) 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 - 7) u d u

Expand

(u - 7) u : u^{\frac{3}{2}} - 7 u

= \frac{1}{4} \cdot \int u^{\frac{3}{2}} - 7 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 7 u d u)

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

\int 7 u d u = \frac{14}{3} u^{\frac{3}{2}}

= \frac{1}{4} (\frac{2}{5} u^{\frac{5}{2}} - \frac{14}{3} u^{\frac{3}{2}})

Substitute back

u = x^{4} + 7

= \frac{1}{4} (\frac{2}{5} (x^{4} + 7)^{\frac{5}{2}} - \frac{14}{3} (x^{4} + 7)^{\frac{3}{2}})

Simplify

\frac{1}{4} (\frac{2}{5} (x^{4} + 7)^{\frac{5}{2}} - \frac{14}{3} (x^{4} + 7)^{\frac{3}{2}}) : \frac{1}{30} (3 x^{4} - 14) (x^{4} + 7)^{\frac{3}{2}}

= \frac{1}{30} (3 x^{4} - 14) (x^{4} + 7)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{30} (3 x^{4} - 14) (x^{4} + 7)^{\frac{3}{2}} + C

\frac{4 x + 4}{3 x ^{\frac{2}{3}}}

\frac{d}{d x} (cos^{6} (3 x^{2} + 4))

3 sin (2)

3 x^{3} + 5 y^{3} = 1

y^{^{'}} - e^{7 x + y} = 0

What is the integral of x^7sqrt(x^4+7) ?
The integral of x^7sqrt(x^4+7) is 1/30 (3x^4-14)(x^4+7)^{3/2}+C