What is the integral of 2be^{2ax} ?

The integral of 2be^{2ax} is be^{2ax} 1/a+C

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integral of 2be^{2ax}

\int 2 b e^{2 a x} d x

b e^{2 a x} \frac{1}{a} + C

Solution steps

\int 2 b e^{2 a x} d x

Take the constant out:

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

= 2 b \cdot \int e^{2 a x} d x

Apply u-substitution

: \int \frac{e ^{u}}{2 a} d u

= 2 b \cdot \int \frac{e ^{u}}{2 a} d u

Take the constant out:

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

= 2 b \frac{1}{2 a} \cdot \int e^{u} d u

Use the common integral:

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

= 2 b \frac{1}{2 a} e^{u}

Substitute back

u = 2 a x

= 2 b \frac{1}{2 a} e^{2 a x}

Simplify

2 b \frac{1}{2 a} e^{2 a x} : b e^{2 a x} \frac{1}{a}

= b e^{2 a x} \frac{1}{a}

Add a constant to the solution

= b e^{2 a x} \frac{1}{a} + C

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