Question

integrate e^x(2^x-x)dx

Answer 1

\[\int\limits_{}^{}e^x(2^x-x)\]

Answer 2

Oh wow, holy crap.

Answer 3

yea that's what I said

Answer 4

Remember that \[
2 = e^{\ln2} \implies 2^x = e^{x\ln 2}
\]

Answer 5

Use distributive property.

Answer 6

then just do each integral separately.

Answer 7

You can use integration by parts. \\[\int\limits\limits_{}^{} a^x = a^x \ln (a)\ \]

Answer 8

can you help me with that?

Answer 9

i mean \[e^x(e^{xln2}-x)\]

Answer 10

use distributive property first

Answer 11

meaning?

Answer 12

multiply out the brackets :P

Answer 13

\[
c(a+b) = ca+cb
\]

Answer 14

oh yes, of course

Answer 15

\[
\int f(x)+g(x)dx = \int f(x)dx + \int g(x)dx
\]

Answer 16

\(
xe^x
\) requires integration by parts. The other one is just chain rule.

Answer 17

I mean u sub.

Answer 18

xe^x requires u sub right?

Answer 19

No it requires integration by parts. You wanna do \(u=x\) and \(dv = e^x\)

Answer 20

oh yea

Answer 21

\[
(uv)' = u'v + uv' \implies \int udv = uv - \int vdu
\]

Answer 22

aren't the both xe^x if you pull that 2 out in front of the integratl sign?

Answer 23

yes i have memorized that formula

Answer 24

No one is just \(e^{x(\ln(2) + 1)}\) the other is \(xe^x\)