What is the integral of (x^7+1)e^{x^8+8x} ?

The integral of (x^7+1)e^{x^8+8x} is 1/8 e^{x^8+8x}+C

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integral of (x^7+1)e^{x^8+8x}

\int (x^{7} + 1) e^{x^{8} + 8 x} d x

\frac{1}{8} e^{x^{8} + 8 x} + C

Solution steps

\int (x^{7} + 1) e^{x^{8} + 8 x} d x

Apply u-substitution

: \int \frac{e ^{u}}{8} d u

= \int \frac{e ^{u}}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{8} e^{u}

Substitute back

u = x^{8} + 8 x

= \frac{1}{8} e^{x^{8} + 8 x}

Add a constant to the solution

= \frac{1}{8} e^{x^{8} + 8 x} + C

\int \frac{( e ^{- x} )}{x} d x

t \frac{d y}{d t} = y + 4 t^{2} sin (3 t)

\frac{\partial}{\partial y} (\frac{x + y ^{2}}{z})

- 2 sin (t) cos (t)

x^{x}, 1

What is the integral of (x^7+1)e^{x^8+8x} ?
The integral of (x^7+1)e^{x^8+8x} is 1/8 e^{x^8+8x}+C