What is the integral of (4x^3-9x^2+16x)sqrt(x^4-3x^3+8x^2) ?

The integral of (4x^3-9x^2+16x)sqrt(x^4-3x^3+8x^2) is 2/3 x^3(x^2-3x+8)^{3/2}+C

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integral of (4x^3-9x^2+16x)sqrt(x^4-3x^3+8x^2)

\int (4 x^{3} - 9 x^{2} + 16 x) x^{4} - 3 x^{3} + 8 x^{2} d x

\frac{2}{3} x^{3} (x^{2} - 3 x + 8)^{\frac{3}{2}} + C

Solution steps

\int (4 x^{3} - 9 x^{2} + 16 x) x^{4} - 3 x^{3} + 8 x^{2} d x

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x^{4} - 3 x^{3} + 8 x^{2}

= \frac{2}{3} (x^{4} - 3 x^{3} + 8 x^{2})^{\frac{3}{2}}

(x^{4} - 3 x^{3} + 8 x^{2})^{\frac{3}{2}} = x^{3} (x^{2} - 3 x + 8)^{\frac{3}{2}}

= \frac{2}{3} x^{3} (x^{2} - 3 x + 8)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} x^{3} (x^{2} - 3 x + 8)^{\frac{3}{2}} + C

\frac{\partial}{\partial x} (e^{5 x} cos (5 y))

\frac{x ( - 4 x ^{2} + 1 + 1 )}{4}

x \to 0 lim (\frac{4 + x - 4 - x}{4 x})

x \to - \frac{π}{2} lim (- tan (0.5 x))

\frac{\partial}{\partial y} (3 x e^{y})

What is the integral of (4x^3-9x^2+16x)sqrt(x^4-3x^3+8x^2) ?
The integral of (4x^3-9x^2+16x)sqrt(x^4-3x^3+8x^2) is 2/3 x^3(x^2-3x+8)^{3/2}+C