Question

Find or evaluate the integral. (Complete the square, if necessary. Remember to use ln |u| where appropriate.)
(8x − 2)/(x^2 + 2x + 2) dx

Answer 1

hey can you help me on this one too

Answer 2

yeah I'll try

Answer 3

okiii

Answer 4

let me know if you need more help

Answer 5

you have a 'typo'

Answer 6

hmm um at the end is that thue answer...

Answer 7

\[x^2 + 2x + 2=x^2+2x+1-1+2\]

\[=(x^2+2x+1)+1=(x+1)^2+1\]

Answer 8

you are right i added 1 twice

Answer 9

(x+1)^2+3 is what myininaya has

Answer 10

I would break it up into two integrals

Answer 11

yes i added 1 twice
its a mistake it should be what zarkon has

Answer 12

so we should
\[2\int\limits_{}^{}\frac{4\tan(\theta)-5}{\sec^2(\theta)} d \theta\]

Answer 13

Yeah Zarkon that's what I was thinking too...
\[\int\limits_{}^{} \frac{4(2x+2)}{x^2+2x+2} \ + \int\limits_{}^{ }\frac{-10}{x^2 + 2x+2} \]

Answer 14

that works nice

Answer 15

\[2(\int\limits_{}^{}\frac{4\tan(\theta)}{\sec^2(\theta)} d \theta-\int\limits_{}^{}\frac{5}{\sec^2(\theta)} d \theta)\]

Answer 16

\[\int\limits_{}^{}2\sin(\theta)\cos(\theta) d \theta-10\int\limits_{}^{}\cos^2(\theta) d \theta\]
\[\int\limits_{}^{}\sin(2\theta) dt heta-10 \int\limits_{}^{}\frac{1}{2}(1+\cos(2\theta))d \theta\]

Answer 17

\[let u = x^2 + 2z + 2; du = (2x + 2)dx \]
So for the left integral we have:
\[4\int\limits_{}^{}\ \frac{du}{u} \ = 4\ln \left| u \right|\]

Answer 18

\[=-\frac{1}{2} \cos(2 \theta)-5(\theta+\frac{1}{2}\sin( 2 \theta))+C\]

Answer 19

now just put it back in terms of x

Answer 20

Right:
\[4\ln \left| x^2 + 2x +2 \right|\]

Answer 21

For the right integral we have to complete the square in the denominator:
\[x^2 + 2x + 2 = (x + 1)^{2} + 1\]

Answer 22

so the final answer is what myininaya wrote out but just put x in place of the theta symbol?

Answer 23

not exactly..it is a little more complicated than that

Answer 24

Therefore the integral will be:\[-10\int\limits_{}^{}\ \frac{dx}{(x+1)^{2}+1} \]

Answer 25

x does not equal theta

Answer 26

\[\tan(\theta)=x+1\]
thats x and theta's relationship

Answer 27

\[\theta=\tan^{-1}(x+1)\]

Answer 28

Now we have to figure out a way to utilize arctan(x) derivative so we must set an intermediate variable like a = x + 1 and if we take the derivative da = dx. So our integral becomes

Answer 29

\[\tan(\theta)=\frac{opposite}{adjacent}=\frac{x+1}{1}=> hyp=\sqrt{(x+1)^2+1}\]
\[\cos(\theta)=\frac{1}{\sqrt{(x+1)^2+1}}, \sin(\theta)=\frac{x+1}{\sqrt{(x+1)^2+1}}\]

Answer 30

\[-10\int\limits_{}^{}\ \frac{da}{a^2 + 1} \ - -10 \tan^{-1} a= -10\tan^{-1} (x+1)\]

Answer 31

Therefore the final answer will be:\[4\ln \left| x^2 + 2x+2 \right| - 10\tan^{-1} (x+1)\]

Answer 32

\[=-\frac{1}{2} \cos(2 \theta)-5(\theta+\frac{1}{2}\sin( 2 \theta))+C \]
\[=\frac{-1}{2}(\cos^2(\theta)-\sin^2(\theta))-5 \theta-\frac{5}{2}2\sin(\theta)\cos(\theta)+C\]
\[=\frac{-1}{2}(\frac{1}{(x+1)^2+1}-\frac{(x+1)^2}{(x+1)^2+1})-5\tan^{-1}(x+1)-5\frac{x+1}{(x+1)^2+1}+C\]

Answer 33

and that can be cleaned up

Answer 34

wait so whos rightis nayeaddo right or you myinaya

Answer 35

go through and check our work and you can come to that conclusion
or you can enter it into wolfram to see who;s answer is right

Answer 36

honestly after that i don't feel like checking my work sorry

Answer 37

lol

Answer 38

I'm telling you...your arms are going to fall off

Answer 39

Yeah myininaya your method is very inricate lol...I mean it involves a lot

Answer 40

based on wolfram nayeaddo has it

Answer 41

http://www.wolframalpha.com/input/?i=integrate%28%288x-2%29%2F%28x^2%2B2x%2B2%29%2Cx%29%3B

Answer 42

use that site to check your work

Answer 43

its awesome

Answer 44

nayeddo i learned a new word
thanks

Answer 45

cool what word?

Answer 46

inricate
thats weird it says i spelled it wrong

Answer 47

yeah there is at before the r

Answer 48

and nice!

Answer 49

i became your fan naye because you won against me this round
but i will win next time

Answer 50

hahaha...I like that competition lol I also became your fan so I guess its on..but like I said your methods are interesting too so we'll see

Answer 51

i guess it would had been better if i separated the fraction at the beginning

Answer 52

or easier

Answer 53

yeah i think especially for the one who asked the question

Answer 54

But it also makes the math easier to understand.. less complicated

Answer 55

but complications make things more dreamy

Answer 56

that's not suppose to make too much sense
so don't read too much in that statement

Answer 57

yeah...I kind of know what you mean though