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

The integral of-6x(-4x^2+6)^5 is 1/8 (-4x^2+6)^6+C

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

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

\frac{1}{8} (- 4 x^{2} + 6)^{6} + C

Solution steps

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

Take the constant out:

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

= - 6 \cdot \int x (- 4 x^{2} + 6)^{5} d x

Apply u-substitution

: \int - \frac{u ^{5}}{8} d u

= - 6 \cdot \int - \frac{u ^{5}}{8} d u

Take the constant out:

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

= - 6 (- \frac{1}{8} \cdot \int u^{5} d u)

Apply the Power Rule

: \frac{u ^{6}}{6}

= - 6 (- \frac{1}{8} \cdot \frac{u ^{6}}{6})

Substitute back

u = - 4 x^{2} + 6

= - 6 (- \frac{1}{8} \cdot \frac{( - 4 x ^{2} + 6 ) ^{6}}{6})

Simplify

- 6 (- \frac{1}{8} \cdot \frac{( - 4 x ^{2} + 6 ) ^{6}}{6}) : \frac{1}{8} (- 4 x^{2} + 6)^{6}

= \frac{1}{8} (- 4 x^{2} + 6)^{6}

Add a constant to the solution

= \frac{1}{8} (- 4 x^{2} + 6)^{6} + C

\int \frac{9}{15 sin ( x ) - 8 cos ( x )} d x

\frac{d}{d x} (\frac{2 x}{a ^{2} - x ^{2}})

n = 1 \sum \infty \frac{1}{4 n ^{2} + 7}

y^{^{''}} + a y = 0, y (0) = 1, y (1) = 0

(8.2) (9.3)

What is the integral of-6x(-4x^2+6)^5 ?
The integral of-6x(-4x^2+6)^5 is 1/8 (-4x^2+6)^6+C