Question

I AM REVISING -THE DEEPER I GO, THE HARDER!
INTEGRATE x(2^x)

Answer 1

\[\int\limits_{}^{}x2^{x}\]

Answer 2

When we have the product of functions in a integral, it's a little hard to solve, because I can't make any substitution or direct solving.
But remembering one thing the derivative of a product, it had this form:
\[(g(x)f(x))'=g'(x)f(x)+g(x)f'(x)\]
now, something what happens if I integrate these?:
\[\int\limits_{}^{} (g(x)f(x))' dx=\int\limits g'(x)f(x) dx + \int\limits g(x)f'(x) dx \]
I'll re-write these:
\[\int\limits g(x)f'(x) dx = \int\limits (g(x)f(x))' dx - \int\limits g'(x)f(x) dx\]
Thre is a very interesting theorem that states this:
\[\int\limits u' dx = u\]
where u is a function and u' it's derivative form. And also:
\[g'(x)=\frac{ dg }{ dx } => g'(x)dx=dg\]
\[f'(x)=\frac{ df }{ dx } => f'(x)dx=df\]
Then, replacing:
\[\int\limits g(x)df=g(x)f(x)- \int\limits f(x)dg\]
you might know this formula this way:
\[\int\limits u dv = uv - \int\limits v du\]
And that is what we call "integral by parts" formula.

Answer 3

ok, let me try to follow here

Answer 4

First try choosing wich one will be "u" and wich one will be "v".

Answer 5

there is a good way of knowing wich one to choose.
it's called ILATE, or in other words: "Inverse, Logarithmic, Algebraic, Trigonometric, Exponential".
"x" is algebraic and "2exp x " is exponential.
So it's conveninient to choose "x" as "u" and "2exp x " as "v"

Answer 6

oh, thank for ILATE

Answer 7

Okay, you can show me how you're doing when you try it out :)

Answer 8

YES

Answer 9

u =x, u' = dx
v' = 2^x dx, v = (2^x)/log 2

Answer 10

BUT THEN it takes me back to \[\int\limits_{}^{} x2^{x}=\]\[x \frac{ 2^{x} }{ \log2 }- \int\limits_{}^{}x\frac{ 2^{x} }{ \log2 }\]

Answer 11

@owlcoffee

Answer 12

@nincompoop

Answer 13

i am stuck, seriously...

Answer 14

how did u get 'x' inside 2nd integral?
isn't du = dx ? and not x

\(\large x \frac{ 2^{x} }{ \log2 }- \int\limits_{}^{}\frac{ 2^{x} }{ \log2 }\color{green}{dx}\)

Answer 15

log2 is a constant, you can pull it out of the integral. and integrate 2 exp x.

I'm getting very rusty in mathematics.

Answer 16

oh, ok, let me correct that

Answer 17

\[\int\limits_{}^{}x2^{x}dx= \frac{x2^{x}}{\log2} -\int\limits_{}^{}\frac{ 2^{x} }{ \log2 }dx\]
=
\[\frac{ x2^{x} }{ \log2 }-\frac{ 2^{x} }{( \log2)^{2} }\]

Answer 18

i think this z ryt

Answer 19

It's By Parts, not magic.

\(\int x\cdot 2^{x}\;dx = \int x\;d\left(\dfrac{2^{x}}{\log(2)}\right) = \dfrac{x\cdot 2^{x}}{\log(2)} - \int\dfrac{2^{x}}{\log(2)}\;dx\)

Answer 20

by parts can be organized in a table: