Question

Can someone help me out with this integral?

Answer 1

\[\int\limits \frac{ 3^2x-2^(x+1) }{ 2^x }\]

Answer 2

wrong, it's integral of {3^(2x)-2^(x+1)}/2^x

Answer 3

who can help me out?! PLEASE

Answer 4

switch the 2^x into e and this could look a lot easier

\[2^{x} = e ^ {\ln 2 ^{x} } = e ^ { x \ln 2}\]

Answer 5

\[I=\int\limits \frac{3^{2x}-(2^{x+1})}{2^x}dx=\int\limits \frac{3^{2^x}-(2^x.2)}{2^x}dx\]
\[2^x=t\]\[2^x \log_{e}2=\frac{dt}{dx}\]\[dx=\frac{1}{\log_{e}2}.\frac{dt}{t}\]\[I=\frac{1}{\log_{e}2}\int\limits \frac{3^t-2t}{t^2}dt\]
See if you can go anywhere from here

Answer 6

\[\int\limits_{}^{} \frac{ 3^{2x} - 2^{x + 1}}{ 2 ^ {x} } = \]
\[\int\limits_{}^{} \frac{ 3 ^{2x} }{ 2 ^{x} } - 2\]

now convert to e's using the conversion given above and the following:
\[3 ^{2x} = e ^{2x \ln 3}\]

Answer 7

thanks allot!