What is the integral of (x^5+1)e^{x^6+6x} ?

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

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integral of (x^5+1)e^{x^6+6x}

\int (x^{5} + 1) e^{x^{6} + 6 x} d x

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

Solution steps

\int (x^{5} + 1) e^{x^{6} + 6 x} 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} + 6 x

= \frac{1}{6} e^{x^{6} + 6 x}

Add a constant to the solution

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

\frac{\partial}{\partial x} (x (x + 1)^{2})

n = 1 \sum \infty \frac{4}{n ^{2} + 5 n + 6}

\int \frac{sin ( \frac{9}{x} )}{x ^{2}} d x

\frac{\partial}{\partial x} (x^{2} - 10 x + y^{2} + 8)

\frac{1}{( s ^{3} + 4 s )}

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