What is the integral of (x^2(3-x))/(sqrt(x^4-4x^3)) ?

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

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integral of (x^2(3-x))/(sqrt(x^4-4x^3))

\int \frac{x ^{2} ( 3 - x )}{x ^{4} - 4 x ^{3}} d x

- \frac{1}{2} x^{4} - 4 x^{3} + C

Solution steps

\int \frac{x ^{2} ( 3 - x )}{x ^{4} - 4 x ^{3}} d x

Apply u-substitution

: \int - \frac{1}{4 u} d u

= \int - \frac{1}{4 u} 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 \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= - \frac{1}{4} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{4} - 4 x^{3}

= - \frac{1}{4} \cdot 2 (x^{4} - 4 x^{3})^{\frac{1}{2}}

Simplify

- \frac{1}{4} \cdot 2 (x^{4} - 4 x^{3})^{\frac{1}{2}} : - \frac{1}{2} x^{4} - 4 x^{3}

= - \frac{1}{2} x^{4} - 4 x^{3}

Add a constant to the solution

= - \frac{1}{2} x^{4} - 4 x^{3} + C

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 3, y^{^{'}} (0) = - 1

\frac{d}{d x} (\frac{e ^{- 2.8 x} - e ^{- 4.6 x}}{1.8})

arcsin (2 x)

\frac{80}{s ^{2} + 40 s}

\frac{d}{d x} (e^{x} + 2^{- x} + 2 cos (x) - 6)

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