What is the integral of x^3sqrt(x^2+6) ?

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

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integral of x^3sqrt(x^2+6)

\int x^{3} x^{2} + 6 d x

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

Solution steps

\int x^{3} x^{2} + 6 d x

Apply u-substitution

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

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

Expand

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

= \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

= \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{u ^{5}}{5} - 2 u^{3}

Substitute back

u = x^{2} + 6

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = \frac{2 x - 1}{y ^{2}}

\frac{\partial}{\partial x} (x y (1 - 7 x - 6 y))

- \frac{( x ^{2} )}{y ^{2}}

\int \frac{( 1 + x ) ^{2}}{x ^{2} + 1} d x

\frac{( 8 u ^{2} )}{( u ^{2} + u ) ^{3}}

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