What is the antiderivative of (4^x - 2^x + 1) / 2^x?
L Luna642 New member Joined Feb 20, 2020 Messages 4 Feb 20, 2020 #1 What is the antiderivative of (4^x - 2^x + 1) / 2^x?
pka Elite Member Joined Jan 29, 2005 Messages 11,971 Feb 20, 2020 #2 Luna642 said: What is the antiderivative of (4^x - 2^x + 1) / 2^x? Click to expand... Write \(\dfrac{4^x-2^x+1}{2^x}\) as \(2^x-1+2^{-x}\). Can you continue? Please post you own work.
Luna642 said: What is the antiderivative of (4^x - 2^x + 1) / 2^x? Click to expand... Write \(\dfrac{4^x-2^x+1}{2^x}\) as \(2^x-1+2^{-x}\). Can you continue? Please post you own work.
Steven G Elite Member Joined Dec 30, 2014 Messages 14,406 Feb 20, 2020 #3 Are you integrating wrt x, y , t ....?
D Deleted member 4993 Guest Feb 20, 2020 #4 Luna642 said: What is the antiderivative of (4^x - 2^x + 1) / 2^x? Click to expand... Do you know the anti-derivative of: 2x.............................. w.r.t. x? Do you know the derivative of: .......................................................added later 2x.............................. w.r.t. x? Last edited by a moderator: Feb 20, 2020
Luna642 said: What is the antiderivative of (4^x - 2^x + 1) / 2^x? Click to expand... Do you know the anti-derivative of: 2x.............................. w.r.t. x? Do you know the derivative of: .......................................................added later 2x.............................. w.r.t. x?
Steven G Elite Member Joined Dec 30, 2014 Messages 14,406 Feb 20, 2020 #5 Let's see. y=ax. ln(y) = xln(a) (1/y)y' = ln(a) Then y' = yln(a) So y' = axln(a) Does that help?