What is the integral of (x^6-3x^2)^4(x^5-x) ?

The integral of (x^6-3x^2)^4(x^5-x) is 1/30 (x^6-3x^2)^5+C

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integral of (x^6-3x^2)^4(x^5-x)

\int (x^{6} - 3 x^{2})^{4} (x^{5} - x) d x

\frac{1}{30} (x^{6} - 3 x^{2})^{5} + C

Solution steps

\int (x^{6} - 3 x^{2})^{4} (x^{5} - x) d x

Apply u-substitution

: \int \frac{u ^{4}}{6} d u

= \int \frac{u ^{4}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int u^{4} d u

Apply the Power Rule

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

= \frac{1}{6} \cdot \frac{u ^{5}}{5}

Substitute back

u = x^{6} - 3 x^{2}

= \frac{1}{6} \cdot \frac{( x ^{6} - 3 x ^{2} ) ^{5}}{5}

Simplify

\frac{1}{6} \cdot \frac{( x ^{6} - 3 x ^{2} ) ^{5}}{5} : \frac{1}{30} (x^{6} - 3 x^{2})^{5}

= \frac{1}{30} (x^{6} - 3 x^{2})^{5}

Add a constant to the solution

= \frac{1}{30} (x^{6} - 3 x^{2})^{5} + C

\frac{d}{d t} (3 - 3 t)

\frac{\partial}{\partial x} (cos (x) y)

5 x, x^{2} - 6

\int_{0}^{2 π} cos (2 x) d x

f (x) = 32 - x

What is the integral of (x^6-3x^2)^4(x^5-x) ?
The integral of (x^6-3x^2)^4(x^5-x) is 1/30 (x^6-3x^2)^5+C