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

The integral of x(x^2-3)^3 is 1/8 (x^2-3)^4+C

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integral of x(x^2-3)^3

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

\frac{1}{8} (x^{2} - 3)^{4} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

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

Substitute back

u = x^{2} - 3

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

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} - 3 ) ^{4}}{4} : \frac{1}{8} (x^{2} - 3)^{4}

= \frac{1}{8} (x^{2} - 3)^{4}

Add a constant to the solution

= \frac{1}{8} (x^{2} - 3)^{4} + C

\int \frac{1}{16 - 9 x ^{2}} d x

\frac{\partial}{\partial x} (a^{- x})

\int_{0}^{π} 8 sin (\frac{x}{2}) d x

y^{^{''}} - 6 y^{^{'}} + 8 y = - 2 cos (4 t)

\int_{0}^{2} 8 t d t

What is the integral of x(x^2-3)^3 ?
The integral of x(x^2-3)^3 is 1/8 (x^2-3)^4+C