What is the integral of x^5sqrt(x^3+7) ?

The integral of x^5sqrt(x^3+7) is 1/3 (2/5 (x^3+7)^{5/2}-14/3 (x^3+7)^{3/2})+C

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integral of x^5sqrt(x^3+7)

\int x^{5} x^{3} + 7 d x

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

Solution steps

\int x^{5} x^{3} + 7 d x

Apply u-substitution

: \int \frac{( u - 7 ) u}{3} d u

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

Expand

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

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

Substitute back

u = x^{3} + 7

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

Add a constant to the solution

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

\frac{1}{( s ^{2} + 1 ) ( s ^{2} + 25 )}

y^{^{'}} = \frac{( 5 x ^{3} + y ^{3} )}{x y ^{2}}

f (x) = 6 x + 1

\frac{d}{d x} (2 3^{x})

r (x) = sin (x^{2}) + \frac{1 - x}{1 + x}

What is the integral of x^5sqrt(x^3+7) ?
The integral of x^5sqrt(x^3+7) is 1/3 (2/5 (x^3+7)^{5/2}-14/3 (x^3+7)^{3/2})+C