Question

Find the integral (1,0) of (x+1)/(2x^2 +4x +3) dx

Answer 1

\[\huge \int\limits_0^1 \frac{x+1}{\color{royalblue}{2x^2+4x+3}}dx\]
Ok this will be a fairly straight forward U-substitution.
Let the blue part be your U.

Answer 2

Did I write the boundaries correctly? 0 is the lower boundary yes?
I just got confused by the way you wrote it.
We generally write the lower number first.
"From 0 to 1" (0,1)

Answer 3

yes

Answer 4

\[\large \color{royalblue}{u=2x^2+4x+3}\]We'll take the derivative of both sides with respect to x,\[\large \dfrac{du}{dx}=4x+4\]We'll move the dx to the other side,\[\large du=4x+4 \;dx\]We'll factor a 4 out from each term on the right,\[\large du=4(x+1) \;dx\]Divide both sides by 4,\[\large \color{royalblue}{\dfrac{1}{4}du=x+1\;dx}\]

Answer 5

Here is a nice way we can rewrite our integral to understand what is going on.\[\large \int\limits\limits_0^1 \frac{\color{orangered}{x+1\;dx}}{\color{royalblue}{2x^2+4x+3}}\]
Woops I colored the du wrong :d let's try that again. So here are the pieces we want to plug into our integral.\[\large \color{royalblue}{u=2x^2+4x+3}\]\[\large \color{orangered}{\dfrac{1}{4}du=x+1\;dx}\]
Understand how they will plug in? :)

Answer 6

The steps we applied to U may have seemed strange, but maybe it makes a little more sense now. We needed to solve for x+1 dx in order to replace it with something in du.
That's why we had to get the 4 factored out and moved away.

Answer 7

So does du get plugged into the bottom?

Answer 8

The blue will replace the blue, and the orange will replace the orange.
So the orange is on top, yes? And that's the one involving du.

Answer 9

so it will be (x+1) dx/du?

Answer 10

and then I plug in (1/4) [f(1)-f(0)]?

Answer 11

\[\large \int\limits\limits\limits_{x=0}^1 \frac{\color{orangered}{x+1\;dx}}{\color{royalblue}{2x^2+4x+3}}\qquad \rightarrow \qquad \int\limits\limits\limits\limits_{x=0}^1 \frac{\color{orangered}{\frac{1}{4}du}}{\color{royalblue}{u}}\]

Answer 12

ok, but then what do I plug in?

Answer 13

The orange turns into the other orange.
The blue turns into the other blue.

Answer 14

From here, the problem becomes a lot easier to integrate.\[\large \dfrac{1}{4}\int\limits\limits\limits\limits\limits_{x=0}^1 \dfrac{du}{u}\]

The only thing to be careful about is the boundaries.
They are still in terms of X while our integral is in terms of U.

We have 2 options.
We can convert the boundaries to U now,
`or`
We can integrate, and then UNDO our U substitution and simply use the boundaries for x.

Answer 15

can you show me how to do both?

Answer 16

If we wanted to convert the boundaries NOW, we would plug the boundaries into our substitution.

\[\large x=0 \qquad \rightarrow \qquad \color{royalblue}{u=2x^2+4x+3} \qquad \rightarrow \qquad \color{royalblue}{u=2(0)^2+4(0)+3}\]So our new lower boundary will be,\[\large \color{royalblue}{u=3}\]

Answer 17

Then plugging x=1 into our substitution gives us,\[\large \color{royalblue}{u=9}\]For the upper limit of integration.

Answer 18

\[\large \dfrac{1}{4}\int\limits\limits\limits\limits\limits\limits_{x=0}^1 \dfrac{du}{u} \qquad \rightarrow \qquad \large \dfrac{1}{4}\int\limits\limits\limits\limits\limits\limits_{u=3}^9 \dfrac{du}{u}\]

Answer 19

ok that makes sense so far

Answer 20

Remember the integral for 1/x? :)
It's a term that we can't apply the Power Rule to since it's that "special" -1 power.

Answer 21

yeah

Answer 22

lol then what is it? XD

Answer 23

(1/4) (1/9) -(1/4)(1/3)?

Answer 24

Woops you plugged in the boundaries, but you forgot to integrate first.

Answer 25

do we convert it back into x terms then?

Answer 26

\[\large (\ln u)'=\dfrac{1}{u}\]Remember this good ole derivative?
We want to do this in reverse.

Answer 27

so (1/4) ln(9) -(1/4) ln(3)

Answer 28

Yes good :)

Answer 29

So there is our answer, you certainly could apply some rules of logs to simplify it down a tiny bit.
But no big deal.


Ok want to see the other method?

Answer 30

so i just solve and that's it?

Answer 31

Yep c:

Answer 32

we submit our answers electronically ...it lets us try problems over and over again...for some reason its saying that 0.275 is wrong

Answer 33

Ok lemme go through the steps on paper real quick to make sure I didn't make a mistake.

It's possible that the computer just wants the answer in a certain format.
We'll figure it out.

Answer 34

ok, and thanks for your help...i really appreciate it

Answer 35

You're probably not suppose to round your answer.
So let's put an exact value in.\[\large \frac{1}{4}\ln9-\frac{1}{4}\ln3\]Factoring out the 1/4 from each term gives us,\[\large \frac{1}{4}\left(\ln9-\ln3\right)\]Then applying a rule of logs gives us,\[\large \frac{1}{4}\ln\left(\dfrac{9}{3}\right) \qquad\rightarrow\qquad \frac{1}{4}\ln3\]

Answer 36

ah ok, that definitely worked. so then regarding all answers with ln...should i just leave them unrounded?

Answer 37

If you're entering these online, then leave them exact unless the directions specify.
It'll usually tell you something like (round your answer to the nearest tenth.) if they want you to round.

Answer 38

ok, will do. thanks for all your help zepdrix :)

Answer 39

np \c:/