What is the integral of x^3sqrt(x^2+12) ?

The integral of x^3sqrt(x^2+12) is 1/5 (x^2+12)^{5/2}-4(x^2+12)^{3/2}+C

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integral of x^3sqrt(x^2+12)

\int x^{3} x^{2} + 12 d x

\frac{1}{5} (x^{2} + 12)^{\frac{5}{2}} - 4 (x^{2} + 12)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 12 d x

Apply u-substitution

: \int (u^{2} - 12) u^{2} d u

= \int (u^{2} - 12) u^{2} d u

Expand

(u^{2} - 12) u^{2} : u^{4} - 12 u^{2}

= \int u^{4} - 12 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 12 u^{2} d u

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

\int 12 u^{2} d u = 4 u^{3}

= \frac{u ^{5}}{5} - 4 u^{3}

Substitute back

u = x^{2} + 12

= \frac{( x ^{2} + 12 ) ^{5}}{5} - 4 (x^{2} + 12)^{3}

Simplify

\frac{( x ^{2} + 12 ) ^{5}}{5} - 4 (x^{2} + 12)^{3} : \frac{1}{5} (x^{2} + 12)^{\frac{5}{2}} - 4 (x^{2} + 12)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 12)^{\frac{5}{2}} - 4 (x^{2} + 12)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 12)^{\frac{5}{2}} - 4 (x^{2} + 12)^{\frac{3}{2}} + C

\frac{\partial}{\partial x} (x y^{2} z^{3})

4 x - x^{2}, 0, 2

x^{2} - 4, - 2, 2

f (x) = (7 x^{3} + x^{2})^{\frac{2}{3}}

\int \frac{1}{( x - 3 ) ( x ^{2} + 1 )} d x

What is the integral of x^3sqrt(x^2+12) ?
The integral of x^3sqrt(x^2+12) is 1/5 (x^2+12)^{5/2}-4(x^2+12)^{3/2}+C