What is the integral of-15x^4(-3x^5-1)^5 ?

The integral of-15x^4(-3x^5-1)^5 is 1/6 (-3x^5-1)^6+C

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integral of-15x^4(-3x^5-1)^5

\int - 15 x^{4} (- 3 x^{5} - 1)^{5} d x

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

Solution steps

\int - 15 x^{4} (- 3 x^{5} - 1)^{5} d x

Take the constant out:

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

= - 15 \cdot \int x^{4} (- 3 x^{5} - 1)^{5} d x

Apply u-substitution

: \int - \frac{u ^{5}}{15} d u

= - 15 \cdot \int - \frac{u ^{5}}{15} d u

Take the constant out:

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

= - 15 (- \frac{1}{15} \cdot \int u^{5} d u)

Apply the Power Rule

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

= - 15 (- \frac{1}{15} \cdot \frac{u ^{6}}{6})

Substitute back

u = - 3 x^{5} - 1

= - 15 (- \frac{1}{15} \cdot \frac{( - 3 x ^{5} - 1 ) ^{6}}{6})

Simplify

- 15 (- \frac{1}{15} \cdot \frac{( - 3 x ^{5} - 1 ) ^{6}}{6}) : \frac{1}{6} (- 3 x^{5} - 1)^{6}

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

Add a constant to the solution

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

\frac{d}{d x} (x ln (4 x))

4 - x, 0, - 2, 6

y = 1, x = y^{\frac{5}{2}}, [0, 1]

\int 6 e^{- 6 x} d x

x \to 1 lim (ln (x) - ln (x - 1))

What is the integral of-15x^4(-3x^5-1)^5 ?
The integral of-15x^4(-3x^5-1)^5 is 1/6 (-3x^5-1)^6+C