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

The integral of x(x^2-3)^4 is 1/10 (x^2-3)^5+C

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

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

\frac{1}{10} (x^{2} - 3)^{5} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{4}}{2} d u

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

Apply the Power Rule

: \frac{u ^{5}}{5}

= \frac{1}{2} \cdot \frac{u ^{5}}{5}

Substitute back

u = x^{2} - 3

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

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} - 3 ) ^{5}}{5} : \frac{1}{10} (x^{2} - 3)^{5}

= \frac{1}{10} (x^{2} - 3)^{5}

Add a constant to the solution

= \frac{1}{10} (x^{2} - 3)^{5} + C

\frac{d}{d x} (- x + 3)

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

2 \frac{d y}{d x} - \frac{( 45 x ^{4} )}{y} = 0

f (x) = x + 4

\frac{x ^{2} - 2 x + 2}{x - 1}

What is the integral of x(x^2-3)^4 ?
The integral of x(x^2-3)^4 is 1/10 (x^2-3)^5+C