Question

integrate x^33^x dx

Answer 1

product of x raised to power three and 3 raised to power x

Answer 2

i assume this is
\[\int x^33^xdx\]

Answer 3

ya

Answer 4

it is a set up for integration by parts, but you have to do it 3 times ! will reduce the power on
\[x^3\] once each time

Answer 5

i have assumed f(x)=x^3 and g'(x)=3^x

Answer 6

for example in the first time you will put
\[u=x^3, dv=3^xdx, du=3x^2,v=\frac{3^x}{\ln(3)}\] and get
\[uv-\int vdu=\frac{x^33^x}{\ln(3)}-\frac{3}{\ln(3)}\int x^23^xdx\]

Answer 7

then lather, rinse, repeat. hold on

Answer 8

easier to read it , click "show steps"
http://www.wolframalpha.com/input/?i=x^3*3^x+dx

Answer 9

how would u integrate 3x^2.3^x/log3

Answer 10

first of all pull out the constants like i did above and get
\[-\frac{3}{\ln(3)}\int x^23^xdx\] then repeat the parts, this time with
\[u=x^2\]

Answer 11

ur ans went over my head

Answer 12

ok it is confusing, but if you got the first one right you should have a good idea what is going on
each time you integrate by parts, you will reduce the power by one when you take the derivative of
\[x^3, x^2, x\]

Answer 13

don't forget that
\[\ln(3)\] is a number (constant) so it come right out front of the integral sign

Answer 14

your next job is really
\[\int x^2 3^xdx\] the constant just sits out front. integrate by parts again

Answer 15

let me show u the first step......3x^2.3x/log3-integrate3x^2.3^x/log3 dx

Answer 16

ah first step is wrong that is the problem

Answer 17

it is
\[\int udv=uv-\int vdu\]

Answer 18

i dnt use that formula

Answer 19

\[u=x^3,du = 3x^2, dv = 3^x, v = \int 3^x=\frac{3^x}{\ln(3)}\]

Answer 20

\[uv=\frac{x^33^x}{\ln(3)}\]

Answer 21

and i stress again, even though the second one is
\[-\int \frac{3x^23^x}{\ln(3)}dx\] you should rewrite it immediately as
\[-\frac{3}{\ln(3)}\int x^23^xdx\]

Answer 22

i m having trouble with the formula implementation......wht to do?