What is the integral of (x^3-1)^4x^2 ?

The integral of (x^3-1)^4x^2 is 1/15 (x^3-1)^5+C

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integral of (x^3-1)^4x^2

\int (x^{3} - 1)^{4} x^{2} d x

\frac{1}{15} (x^{3} - 1)^{5} + C

Solution steps

\int (x^{3} - 1)^{4} x^{2} d x

Apply u-substitution

: \int \frac{u ^{4}}{3} d u

= \int \frac{u ^{4}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int u^{4} d u

Apply the Power Rule

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

= \frac{1}{3} \cdot \frac{u ^{5}}{5}

Substitute back

u = x^{3} - 1

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

Simplify

\frac{1}{3} \cdot \frac{( x ^{3} - 1 ) ^{5}}{5} : \frac{1}{15} (x^{3} - 1)^{5}

= \frac{1}{15} (x^{3} - 1)^{5}

Add a constant to the solution

= \frac{1}{15} (x^{3} - 1)^{5} + C

\int \frac{1}{u ^{2} + a ^{2}} d u

n = 1 \sum \infty \frac{1}{n ( n + 2 )}

1 - 4 x

n = 1 \sum \infty \frac{( 1 )}{n ^{2}}

\frac{d ^{2}}{d x ^{2}} (ln (1 + e^{x}))

What is the integral of (x^3-1)^4x^2 ?
The integral of (x^3-1)^4x^2 is 1/15 (x^3-1)^5+C