What is the integral of e^{9x+e^{9x}} ?

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

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integral of e^{9x+e^{9x}}

\int e^{9 x + e^{9 x}} d x

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

Solution steps

\int e^{9 x + e^{9 x}} d x

Apply u-substitution

: \int e^{u + e^{u}} \frac{1}{9} d u

= \int e^{u + e^{u}} \frac{1}{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 + e^{u}} d u

Apply u-substitution

: \int e^{v} d v

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

Use the common integral:

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

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

Substitute back

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

Add a constant to the solution

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

\int (e^{x} + 2) (x - 5) d x

\int \frac{sin ( 2 x )}{20 + cos ^{2} ( x )} d x

y = ln (x^{3} sin^{2} (x))

\int x^{7} x^{4} + 2 d x

\frac{1}{x + 5}

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