What is the integral of 3e^{5x+e^{5x}} ?

The integral of 3e^{5x+e^{5x}} is 3/5 e^{e^{5x}}+C

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integral of 3e^{5x+e^{5x}}

\int 3 e^{5 x + e^{5 x}} d x

\frac{3}{5} e^{e^{5 x}} + C

Solution steps

\int 3 e^{5 x + e^{5 x}} d x

Take the constant out:

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

= 3 \cdot \int e^{5 x + e^{5 x}} d x

Apply u-substitution

: \int e^{u + e^{u}} \frac{1}{5} d u

= 3 \cdot \int e^{u + e^{u}} \frac{1}{5} d u

Take the constant out:

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

= 3 \cdot \frac{1}{5} \cdot \int e^{u + e^{u}} d u

Apply u-substitution

: \int e^{v} d v

= 3 \cdot \frac{1}{5} \cdot \int e^{v} d v

Use the common integral:

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

= 3 \cdot \frac{1}{5} e^{v}

Substitute back

= 3 \cdot \frac{1}{5} e^{e^{5 x}}

Simplify

3 \cdot \frac{1}{5} e^{e^{5 x}} : \frac{3}{5} e^{e^{5 x}}

= \frac{3}{5} e^{e^{5 x}}

Add a constant to the solution

= \frac{3}{5} e^{e^{5 x}} + C

\frac{d y}{d x} = \frac{3 x ^{2}}{1 - y}

x^{2} e^{- 5 x}

\int_{\frac{1}{3}}^{5} 6 x ln (3 x) d x

x^{2} \frac{d y}{d x} = y - x y, y (- 1) = - 1

ln (x), at 1

What is the integral of 3e^{5x+e^{5x}} ?
The integral of 3e^{5x+e^{5x}} is 3/5 e^{e^{5x}}+C