Question

∫x^2 3^x dx
u = x^2
du = 2x dx

dv = 3^x
v = 3^x/ ln (3) + c


∫udv = uv - ∫vdu

so

Answer 1

x^2(3^x/ln(3) +c) - ∫ 3^x/ ln(3) +c * x^2
?

Answer 2

your trying to find the integral right?

Answer 3

integrate it by parts.

Answer 4

shamil, you've got the right idea, but when you go from dv to v, we don't need a constant of integration.

Answer 5

Yea, just add the C at the very end, and here you just need to integrate by parts 2ce.

Answer 6

x^2(3^x/ln(3))- ∫ 3^x/ ln(3) * x^2 + C
then?

Answer 7

Woops you plugged in `u` into your second part.
You wanted to plug in `du`, right?

Answer 8

oh oops

Answer 9

x^2(3^x/ln(3))- ∫ 3^x/ ln(3) * 2x + C

Answer 10

Ok yes, looks good now :)\[\Large\bf\frac{1}{\ln3}x^2 3^x-\frac{2}{\ln3}\int\limits x3^x\;dx\]

Answer 11

Thank you!

Answer 12

So you brought x^2 down to x, just needs parts again I guess right? :o

Answer 13

Yeah, i was told i would need to integrate it twice.. not sure about that part tho.

Answer 14

i just do ∫x3^x dx now?

Answer 15

yah, just make sure you `distribute` the \(\Large\bf -\dfrac{2}{ln3}\) to each part of your new setup.

Answer 16

Alright. so
u = x
du = 1
dv = 3^x
v = 3^x/ln(3) + c
?

Answer 17

v=3^x/ln(3)

Answer 18

crapp my game is starting :c

Answer 19

hold up lol
∫udv = uv - ∫vdu
-2/ln3(x3^x - ∫3^x/ln(3) + C)

Answer 20

\[\large\bf\frac{1}{\ln3}x^2 3^x-\frac{2}{\ln3}\left(\color{red}{\frac{1}{\ln3}}x3^x-\frac{1}{\ln3}\int\limits 3^x\;dx\right)\]So the brackets is our second by-parts.
You missed this small piece in red.

Answer 21

oh dang it, but thanks anyways , has been a good learning process