Question

Some integral practice problem....

Answer 1

\[\int\limits_{ }^{ } \frac{1}{2x^2-16x+26}~dx\]Don't do my problem though, giude me if I am stuck...

Answer 2

please note that:
\[\frac{ 1 }{ 2x ^{2}-16x+26 }=\frac{ 1 }{ 2(x ^{2}-8x+16-16+13) }\]

Answer 3

Dang it, I lost my latex.

Answer 4

ohh, wait, -2/

Answer 5

sorry I think:
\[\frac{ 1 }{ 2 }\frac{ 1 }{ (x-4)^{2}-3 }\]

Answer 6

I was so off....
\[\int\limits_{ }^{ } \frac{1}{2(x^2-8x+16)-6}~du\]\[\int\limits_{ }^{ } \frac{1}{2(x-4)^2-6}~du\]

Answer 7

\[\frac{1}{2}\int\limits_{ }^{ } \frac{1}{(x-4)^2-6}~du\]

Answer 8

that's right!

Answer 9

then u=x-4, du=dx

Answer 10

Oh the dus should be dx.

Answer 11

please, wait a moment, I think better is:

Answer 12

\[\frac{1}{2}\int\limits_{ }^{ } \frac{1}{u^2-6}~du\]
like this

Answer 13

\[\frac{ 1 }{ 6 }\frac{ 1 }{ \left( \frac{ x-4 }{ \sqrt{3} } \right)^{2}-1 }\]

Answer 14

I am not a good follower I guess...

Answer 15

sorry I've made an error before, please note that:
\[\frac{ 1 }{ 2[(x-4)^{2}-3] }\]
do you agree?

Answer 16

@idku

Answer 17

well that is what I said...

Answer 18

my last reply in equation editor.

Answer 19

\[\frac{1}{2}\int\limits_{ }^{ } \frac{1}{u^2-6}~du\]

Answer 20

I subed u.

Answer 21

ok! now I write this:
\[\frac{ 1 }{ 2*3[(\frac{ x-4 }{ \sqrt{3} })^{2}-1] }\]
do you agree?

Answer 22

I see the x-4, but it is kind of hard to follow the who thing with the sqrt3/

Answer 23

the whol*le thing...

Answer 24

please I rewrite simplier

Answer 25

I only meant I didn't understand how you obtained the result, nut that it is inefficient or something.

Answer 26

\[\frac{ 1 }{ 2*3[ \frac{ (x-4)^{2} }{ 3 }-1 ]}\]
now?

Answer 27

nvm, I am sorry for wasting your time, I think I'll try to start over from
\[\frac{1}{2}\int\limits_{ }^{ } \frac{1}{x^2-8x+13}~dx\]

Answer 28

I am very bad to follow the skipped steps.

Answer 29

so?

Answer 30

I know another method!

Answer 31

The message I am trying to deliver, is that I just simply don't understand what you are doing.

Answer 32

So, whatever, don't waste your time on me...

Answer 33

ok! nevertheless if you need help, please send me a message!

Answer 34

\[\int\limits_{ }^{ } \frac{1}{2x^2-16x+26}~dx\]\[\frac{1}{2} \int\limits_{ }^{ } \frac{1}{x^2-8x+13}~dx\]

Answer 35

\[\frac{1}{2} \int\limits_{ }^{ } \frac{1}{(x-4)^2-3}~dx\]Ohh,, I see, you divided by 3 on the denominator. Don't know why you experienced a difficulty saying so, but I get it ty.
\[\frac{1}{2} \int\limits_{ }^{ } \frac{1}{3[\frac{(x-4)^2}{3}-1]}~dx\]and then, you distributed the power

Answer 36

\[\frac{1}{2} \int\limits_{ }^{ } \frac{1}{3[\frac{(x-4)^2}{\sqrt{3}^2}-1]}~dx\]\[\frac{1}{2\sqrt{3}} \int\limits_{ }^{ } \frac{1}{[\frac{(x-4)^2}{\sqrt{3}^2}-1]}~dx\]\[\frac{1}{2\sqrt{3}} \int\limits_{ }^{ } \frac{1}{ \left(\begin{matrix} \frac{x-4}{3} \\ \end{matrix}\right) ^2 -1}~dx\]

Answer 37

\[\frac{1}{2\sqrt{3}} \int\limits_{ }^{ } \frac{1}{ \left(\begin{matrix} \frac{x-4}{3} \\ \end{matrix}\right) ^2 -1^2}~dx\]

Answer 38

I think:
\[\frac{ 1 }{ 3 }\]
not:
\[\frac{ 1 }{ 2\sqrt{3} }\]
please check

Answer 39

ohh, true
So it would become
\[\frac{1}{6} \int\limits_{ }^{ } \frac{1}{ \left(\begin{matrix} \frac{x-4}{3} \\ \end{matrix}\right) ^2 -1^2}~dx\]because we already had the 1/2 on te outside.

Answer 40

\[u=\frac{x-4}{3}~~~~~du=dx\]

Answer 41

\[\frac{1}{6} \int\limits_{ }^{ } \frac{1}{u^2 -1}~du\]

Answer 42

furthermore I think:
\[\frac{ x-4 }{ \sqrt{3} }\]
not
\[\frac{ x-4 }{ 3 }\]
please check

Answer 43

yes, sqrt{3}, so I'll sub back the sqrt 3, but as it is now, I will just coontinue... alright?

Answer 44

tnx for catching that

Answer 45

please change variable, namely:
\[t=\frac{ x-4 }{ \sqrt{3} }\]
where t is the new variable

Answer 46

\[\frac{1}{6} \int\limits_{ }^{ } \frac{1}{t^2 -1}~dt\]\[\frac{1}{6} \int\limits_{ }^{ } \frac{1}{(t+1)(t-1)}~dt\]\[\frac{1}{6} \int\limits_{ }^{ } \frac{A}{(t+1)}+\frac{B}{(t-1)}~dt\]not sure why, t, but whatever makes you happy.
\[A(t-1)+B(t+1)=1~~~~~~~~A=-1/2,~~~~B=1/2\]\[\frac{1}{6} \int\limits_{ }^{ } -\frac{1}{2(t+1)}+\frac{1}{2(t-1)}~dt\]

Answer 47

that's right!

Answer 48

\[\frac{1}{6} \int\limits_{ }^{ } \frac{1}{2(t+1)}-\frac{1}{2(t-1)}~dt\]\[\frac{1}{6}\ln \left| 2(t+1) \right|-\frac{1}{6}\ln \left| 2(t-1) \right|+C\]\[\frac{1}{6}\ln \left| (t+1) \right|-\frac{1}{6}\ln \left| (t-1) \right|+C\]removing 1/6(ln(2) because they cancel, and even if they don't they get absorbed by the +C.

Answer 49

\[\frac{1}{6}\ln \left| (\frac{x-4}{\sqrt{3}}+1) \right|-\frac{1}{6}\ln \left| (\frac{x-4}{\sqrt{3}}-1) \right|+C\]

Answer 50

\[\frac{1}{6}\ln \left| (\frac{x-4+\sqrt{3}}{\sqrt{3}}) \right|-\frac{1}{6}\ln \left| (\frac{x-4-\sqrt{3}}{\sqrt{3}}) \right|+C\]

Answer 51

the square roots of 3, are also gone,
\[\frac{1}{6}\ln \left| (x-4+\sqrt{3}) \right|-\frac{1}{6}\ln \left| (x-4-\sqrt{3}) \right|+C\]

Answer 52

please, I think there is a error:
\[\frac{ 1 }{ 12 }\left( \ln (t-1)-\ln(t+1) \right)\]

Answer 53

So, it would be,
\[\frac{1}{6}\ln \left| x-4+\sqrt{3} \right|-\frac{1}{6}\ln \left| x-4-\sqrt{3} \right|+C\]

Answer 54

Wait, why 1/12?

Answer 55

I wasn;t factoring the 2 out of the LNs, I just used the ln(a*b)=ln(a)+ln(b)

Answer 56

because you can factor out 1/2

Answer 57

ohh, I should have done that before when integrated to get natural logs.. I see.
because with 2(t+1) or -,
the dt=2dx, and that would either require\[\frac{1}{12}\ln \left| x-4+\sqrt{3} \right|-\frac{1}{12}\ln \left| x-4-\sqrt{3} \right|+C\] sub or to factor, so it is now supposed to be

Answer 58

I inserted the answer in a wrong place...

Answer 59

t is supposed to be at the ens of the reply.

Answer 60

I got:
\[\frac{ 1 }{ 12 }\ln \left| \frac{ x-4-\sqrt{3} }{ x-4+\sqrt{3} } \right|+C\]

Answer 61

same...

Answer 62

ty

Answer 63

sorry the signs are different

Answer 64

@idku

Answer 65

yeah,, my bad....

Answer 66

@idku Thank you!