E EmmaC New member Joined Jan 22, 2011 Messages 3 Jan 23, 2011 #1 Why does the integration of f(x)= 4x^3+3lnx g(x)= 2(e^x-2) equal x^4 + 3xlnx - 3x + 2e^x-4x + C ? I can't understand where the 3xlnx - 3x comes from? =S Regards, Emma
Why does the integration of f(x)= 4x^3+3lnx g(x)= 2(e^x-2) equal x^4 + 3xlnx - 3x + 2e^x-4x + C ? I can't understand where the 3xlnx - 3x comes from? =S Regards, Emma
A Aladdin Full Member Joined Mar 27, 2009 Messages 551 Jan 23, 2011 #2 Your typing needs a bit of clarification. You've given two functions ; f(x) = 4x^3+3lnx & g(x) = 2(e^x) - 4 . . . Actually, the integration of ln(x) dx = xlnx - x " Use integration by parts " .... Now you can continue ...
Your typing needs a bit of clarification. You've given two functions ; f(x) = 4x^3+3lnx & g(x) = 2(e^x) - 4 . . . Actually, the integration of ln(x) dx = xlnx - x " Use integration by parts " .... Now you can continue ...
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Jan 23, 2011 #3 Hello, EmmaC! It took a while to "read" your question . . . \(\displaystyle \text{Why does: }\;\int\left(4x^3+3\ln x + 2e^x-4)\,dx \;\;=\;\;x^4 + 3x\ln x - 3x + 2e^x-4x + C\, ?\) \(\displaystyle \text{I can't understand where the }3x\ln x - 3x\text{ comes from.}\) Click to expand... Well, what did YOU get for the second integral?
Hello, EmmaC! It took a while to "read" your question . . . \(\displaystyle \text{Why does: }\;\int\left(4x^3+3\ln x + 2e^x-4)\,dx \;\;=\;\;x^4 + 3x\ln x - 3x + 2e^x-4x + C\, ?\) \(\displaystyle \text{I can't understand where the }3x\ln x - 3x\text{ comes from.}\) Click to expand... Well, what did YOU get for the second integral?