How do you integrate #f(x)=5^x/2^x# using the quotient rule?

Question

Isaiah Allen · Accepted Answer

#intf(x)dx=5^x/(2^xln(5/2))+C.#

In the first instance, there is no Rule called Quotient Rule for

Integration !

However, the fun. #f(x)=5^x/2^x#, can easily be integrated, using the

following Standard Integral :-

#inta^xdx=a^x/lna+C, a in RR^+ -{1}, x in RR#.

Since, #5^x/2^x=(5/2)^x, we have,

#intf(x)dx=int(5/2)^xdx=(5/2)^x/ln(5/2)=5^x/(2^xln(5/2))+C.#

Charles Arocha · Answer

undefined To integrate $ f(x) = \frac{5^x}{2^x} $ using the quotient rule, follow these steps:

1. Let $ u(x) = 5^x $ and $ v(x) = 2^x $.
2. Compute the derivatives: $ u'(x) = \ln(5) \cdot 5^x $ and $ v'(x) = \ln(2) \cdot 2^x $.
3. Apply the quotient rule: $ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x) \cdot v(x) - u(x) \cdot v'(x)}{[v(x)]^2} $.
4. Substitute the derivatives and functions into the quotient rule formula.
5. Simplify the expression.

Therefore, the integral of $ f(x) = \frac{5^x}{2^x} $ using the quotient rule is $ \int f(x) \, dx = \frac{\ln(5) \cdot 5^x \cdot 2^x - 5^x \cdot \ln(2) \cdot 2^x}{[2^x]^2} + C $, where $ C $ is the constant of integration.