What is the integral of e^{-x^6}(-6x^5) ?

The integral of e^{-x^6}(-6x^5) is e^{-x^6}+C

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integral of e^{-x^6}(-6x^5)

\int e^{- x^{6}} (- 6 x^{5}) d x

e^{- x^{6}} + C

Solution steps

\int e^{- x^{6}} (- 6 x^{5}) d x

Simplify

= \int - 6 e^{- x^{6}} x^{5} d x

Take the constant out:

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

= - 6 \cdot \int e^{- x^{6}} x^{5} d x

Apply u-substitution

: \int - \frac{e ^{u}}{6} d u

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

Take the constant out:

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

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

Use the common integral:

\int e^{u} d u = e^{u}

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

Substitute back

u = - x^{6}

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

Simplify

- 6 (- \frac{1}{6} e^{- x^{6}}) : e^{- x^{6}}

= e^{- x^{6}}

Add a constant to the solution

= e^{- x^{6}} + C

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

\frac{d y}{d x}, - x + y - x^{3} = y^{3} + 4 x^{2}

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

\int \frac{1}{n ( n + 1 )}

y^{^{''}} - 3 y^{^{'}} - 18 y = 0

What is the integral of e^{-x^6}(-6x^5) ?
The integral of e^{-x^6}(-6x^5) is e^{-x^6}+C