What is the integral of (x^8+1)e^{x^9+9x} ?

The integral of (x^8+1)e^{x^9+9x} is 1/9 e^{x^9+9x}+C

Popular >

integral of (x^8+1)e^{x^9+9x}

\int (x^{8} + 1) e^{x^{9} + 9 x} d x

\frac{1}{9} e^{x^{9} + 9 x} + C

Solution steps

\int (x^{8} + 1) e^{x^{9} + 9 x} d x

Apply u-substitution

: \int \frac{e ^{u}}{9} d u

= \int \frac{e ^{u}}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{1}{9} e^{u}

Substitute back

u = x^{9} + 9 x

= \frac{1}{9} e^{x^{9} + 9 x}

Add a constant to the solution

= \frac{1}{9} e^{x^{9} + 9 x} + C

y^{^{''}} + 19 y = 0

x \to - 5 - lim (\frac{∣ 5 - x ∣}{5 - x})

\int \frac{x ^{2} + 36 x - 9}{x ^{3} - 9 x} d x

\int_{0}^{\frac{π}{2}} tan^{2} (x) d x

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

What is the integral of (x^8+1)e^{x^9+9x} ?
The integral of (x^8+1)e^{x^9+9x} is 1/9 e^{x^9+9x}+C