What is the integral of x^5e^{(x^6-4)} ?

The integral of x^5e^{(x^6-4)} is 1/6 e^{x^6-4}+C

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integral of x^5e^{(x^6-4)}

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

\frac{1}{6} e^{x^{6} - 4} + C

Solution steps

\int x^{5} e^{x^{6} - 4} d x

Apply u-substitution

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

= \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

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

Use the common integral:

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

= \frac{1}{6} e^{u}

Substitute back

u = x^{6} - 4

= \frac{1}{6} e^{x^{6} - 4}

Add a constant to the solution

= \frac{1}{6} e^{x^{6} - 4} + C

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

\frac{\partial}{\partial x} (\frac{x y}{2})

f (x) = 5 x - 4 x, (1, 1)

sin (x) y^{^{'}} + cos (x) y = 0

\frac{9 s}{( s + 9 ) ^{2}}

What is the integral of x^5e^{(x^6-4)} ?
The integral of x^5e^{(x^6-4)} is 1/6 e^{x^6-4}+C