Question

Please Help me :/

Answer 1

\[\int\limits \frac{ 19xe^{2x} }{ (1+2x)^2 } dx\]

Answer 2

Start by substituting \( u = 1+2x \) and taking out all the constants. Then, you should be able to separate it into two integrals that can be evaluated using integration-by-parts.

If you need more clarification, I can type out the first couple of steps.

Answer 3

@andrewyates Could you do so please? I am a bit lost.

Answer 4

\[ = 19\int \frac{\frac{u - 1}{2} e^{u - 1}}{u^2} \frac{du}{2}\] (by substituting \(u=1 + 2x\)).
\[ = \frac{19}{4e} \int \frac{(u-1)e^u}{u^2} du\]

All I'm doing is replacing \(x\) with \(\frac{u-1}{2}\) and using \(\frac{du}{dx} = 2\). Does this make sense?

Answer 5

How did you get \[e^{u-1}\]

Answer 6

\[\large\rm u=1+2x\qquad\qquad\to\qquad\qquad u-1=2x\]

Answer 7

Oh right... i see.
But how did he pull out a 1/4e?

Answer 8

\[= 19\int\limits \frac{\frac{u - 1}{\color{orangered}{2}} e^{u - 1}}{u^2} \frac{du}{\color{orangered}{2}}\]Here is the fourth, do you see it?

Answer 9

Oh. But wouldn't those 2's cancel because one is in the denominator and the other is in the numerator?

Answer 10

No.
You can either apply your silly fraction rule "keep change flip" if you need to,\[\large\rm \frac{\left(\frac12\right)}{\frac21}=\frac{1}{2}\cdot\frac12\]But you should try to get comfortable with fraction stuff like this,\[\large\rm \frac{\left(\frac12\right)}{2}=\frac{1}{2\cdot2}\]

Answer 11

O. okay I see

Answer 12

\[\large\rm e^{u-1}=\frac{e^u}{e^1}=\frac1e\cdot e^u\]So that's where the e is coming from outside of the integral.
He wanted to clean up the exponent a little bit.

Answer 13

Oh okay.

Answer 14

And so before taking the integral, I have distribute the e^u

Answer 15

Mmmm ya that's probably a good idea.

Answer 16

Hm... But how would i integrate that?
\[\frac{ 19 }{ 4 } \int\limits \frac{ ue^{2}-e^u }{ u^2 }du\]

Answer 17

Oops. disregard that 2, i meant u

Answer 18

Can i split it like...\[\frac{ 19 }{ 4 } \left[ \int\limits (ue^u-e^u) -\int\limits \frac{ 1 }{ u^2 } \right]\]

Answer 19

No.
Hmm thinking.

Answer 20

Oh mkay

Answer 21

My first thought was to separate it into two integrals like this,\[\large\rm \frac{19}{4e}\int\limits\frac{e^{u}}{u}du~-\frac{19}{4e}\int\limits\frac{e^u}{u^2}du\]But those won't integrate nicely. hmm

Answer 22

Can we use integation by parts to solve this instead?

Answer 23

Wolfram says that this \(\large\rm \int\frac{(u-1)e^u}{u^2}du\) equals this \(\large\rm \frac{e^u}{u}\)

So it's some very simple integral,
it's just not clicking in my brain ahhh >.<

Answer 24

Ugh, I don't know. I'm stuck.

Answer 25

This u-sub is giving me some trouble...
Going back to the original expression and applying Integration-By-Parts right away seems to be giving me more success.

Answer 26

Hope that helps.
Integrate the exponential, then get a common denominator to simplify.

Answer 27

Oh I made a mistake lemme reupload >.<

Answer 28

So the answer is \[\frac{ -19xe^{2x} }{ 2+4x }-\frac{ 19e^{2x} }{ 4 }\] ?

Answer 29

Hm... It says that my answer is wrong. :/

Answer 30

No, the second term is positive, ya?

Answer 31

Oh yea, let me try it by fixing that.

Answer 32

Oh yes. It worked...... Hehe
Thanks again!

Answer 33

yayyy

Answer 34

did u get it