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7/x - 2e^x Find an anti-derivative of f(x) I almost have it correct I think but I'm messing something up. I moved x to the top making it 7x^-1 - 2e^x but I am unsure how take the derivative of -2e^x
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wolframalpha my friend
\[7\ln x - 2e ^{x} + c\]
thank you! how did you know to to change 7 into a natural log?
dont write 1/x as x^-1..... instead remember the anti derivative of 1/x is \[\ln x\].. and the anti derivative of \[e ^{x}\] is \[e ^{x}\] itself...
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i didn't change 7 into natural log....it was 1/x which got transformed to lnx through integration...
I would write the antiderivative of \(\displaystyle \frac{1}{x}\) as \(\displaystyle\ln|x|+c\)
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