Question

what about this..ive try the substituting method but then stucking...

Answer 1

\[\int\limits_{}^{} x ^{2}e ^{4x} dx\]

Answer 2

What other methods have you tried?

Answer 3

go for integration by parts !!

Answer 4

Seconded ^ :)
\[\Large \int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx\]
\[\Large \int udv=uv-\int vdu\]

Answer 5

@Ika_Kim
You know integration by parts?

Answer 6

urmm...i dont know integration by part....so is that the method that u mention?

Answer 7

Yeah...

Answer 8

sorry the intrnet connection here is very bad.

Answer 9

No worries :)
Remember the product rule?
\[\Large f'(x)g(x)+f(x)g'(x)=\frac{d}{dx}[f(x)g(x)]\]

Answer 10

So, we could multiply everything by dx
\[\Large f'(x)g(x)dx+f(x)g'(x)=d[f(x)g(x)]\]and integrate both sides
\[\Large\int f'(x)g(x)dx+\int f(x)g'(x)=\int d[f(x)g(x)]\]
The right side simplifies to
\[\Large\int f'(x)g(x)dx+\int f(x)g'(x)=\int f(x)g(x)\]
Rearranging gives us
\[\Large\int f(x)g'(x)=f(x)g(x)-\int f'(x)g(x)dx\]Thus the formula for integration by parts :)

Answer 11

Sorry
\[\Large\int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx\]

Answer 12

ok . i see. let me try :D THANKS :d

Answer 13

But it's best to stick to this nicer looking
\[\huge \int udv = uv - \int v du\]

Answer 14

\[x ^{2}(\frac{ e ^{4x} }{ 4 })-\int\limits_{}^{}\frac{ e^{4x} }{ 4 }2xdx\]

Answer 15

@PeterPan what about this ?

Answer 16

btw thans u guys :)

Answer 17

You forgot
\[\large x ^{2}(\frac{ e ^{4x} }{ 4 })-\int\limits_{}^{}\frac{ 2xe^{4x} }{ 4 }2xdx\]

Answer 18

And now, apply Integration by Parts again :)

Answer 19

heyyy where that 2x come from ????

Answer 20

f'(x) :P

Answer 21

Oh... right, I didn't see it, sorry, you were right :)

Answer 22

\[\Large x ^{2}(\frac{ e ^{4x} }{ 4 })-\int\limits_{}^{}\frac{ e^{4x} }{ 4 }2xdx\]

Answer 23

ok ^^ then \[x ^{2}(\frac{ e^{4x} }{ 4 })-\int\limits_{}^{}\frac{ e ^{4x} }{ 2 }xdx\]

Answer 24

mhmm. Now just use Integration by parts on that integral again :)

Answer 25

whats next ???

Answer 26

\[\Large x ^{2}(\frac{ e^{4x} }{ 4 })-\frac12\int\limits_{}^{}xe^{4x}dx\]

Answer 27

Do Integration by Parts on this part...
\[\Large x ^{2}(\frac{ e^{4x} }{ 4 })-\frac12\color{green}{\int\limits_{}^{}xe^{4x}dx}\]

Answer 28

ok...then , after that ,we r done ?

Answer 29

Yeah, after that, one more integral, though, thankfully, no longer by parts :)
Just a normal integral after that.

Answer 30

ohhh..thanks a lot peterpan !!!! :D

Answer 31

No problem :)