What is the integral of e^{2x}5^{2x} ?

The integral of e^{2x}5^{2x} is 1/(2(1+ln(5)))*25^{((1+ln(5))x)/(ln(5))}+C

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integral of e^{2x}5^{2x}

\int e^{2 x} 5^{2 x} d x

\frac{1}{2 ( 1 + ln ( 5 ) )} \cdot 2 5^{\frac{( 1 + l n ( 5 ) ) x}{l n ( 5 )}} + C

Solution steps

\int e^{2 x} \cdot 5^{2 x} d x

5^{2 x} = 2 5^{x}

= \int 2 5^{x} e^{2 x} d x

Apply u-substitution

= \int \frac{u ^{\frac{1}{l n ( 5 )}}}{2 ln ( 5 )} d u

Take the constant out:

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

= \frac{1}{2 ln ( 5 )} \cdot \int u^{\frac{1}{l n ( 5 )}} d u

Apply the Power Rule

= \frac{1}{2 ln ( 5 )} \cdot \frac{1}{1 + ln ( 5 )} ln (5) u^{\frac{1 + l n ( 5 )}{l n ( 5 )}}

Substitute back

u = 2 5^{x}

= \frac{1}{2 ln ( 5 )} \cdot \frac{1}{1 + ln ( 5 )} ln (5) (2 5^{x})^{\frac{1 + l n ( 5 )}{l n ( 5 )}}

Simplify

\frac{1}{2 ln ( 5 )} \cdot \frac{1}{1 + ln ( 5 )} ln (5) (2 5^{x})^{\frac{1 + l n ( 5 )}{l n ( 5 )}} : \frac{1}{2 ( 1 + ln ( 5 ) )} \cdot 2 5^{\frac{( 1 + l n ( 5 ) ) x}{l n ( 5 )}}

= \frac{1}{2 ( 1 + ln ( 5 ) )} \cdot 2 5^{\frac{( 1 + l n ( 5 ) ) x}{l n ( 5 )}}

Add a constant to the solution

= \frac{1}{2 ( 1 + ln ( 5 ) )} \cdot 2 5^{\frac{( 1 + l n ( 5 ) ) x}{l n ( 5 )}} + C

What is the integral of e^{2x}5^{2x} ?
The integral of e^{2x}5^{2x} is 1/(2(1+ln(5)))*25^{((1+ln(5))x)/(ln(5))}+C