Question

I am practicing another integral. (check, but don't figure out what I should see for me)

Answer 1

\[\large \int\limits_{ }^{ } \frac{x-1}{x^2+2}~dx\]

Answer 2

wait, can I get a hint, but just a hint...

Answer 3

Hint: u-substitution

Answer 4

I was thinking of it, but which sub for what, for x-1?

Answer 5

well, lets go that way.

Answer 6

\[\large \int\limits_{ }^{ } \frac{u}{u^2-2x+3}~du\]

Answer 7

Do you like what you see?

Answer 8

well, not really, but seems a bit better than before.

Answer 9

hold on, I can factor

Answer 10

No I can't

Answer 11

ohh, SHhhh the x, it is u

Answer 12

Hint : split the given integral into two integrals, one can be worked by simple u substitution and the other has an obvious trig substitution

Answer 13

\[\large \int\limits_{ }^{ } \frac{x}{x^2+2}~dx-\int\limits_{ }^{ } \frac{1}{x^2+2}dx\]

Answer 14

yes, the first one:
u=x^2, (1/2)du=x dx

Answer 15

Ohh, u=x62+2

Answer 16

Yes, i thought of that

Answer 17

yes that substitution gives u a simpler integrand
you need to look ahead of what you get by ur substition

integration requires guessing

Answer 18

\[\frac{1}{2}\large \int\limits_{ }^{ } \frac{1}{u}~du-\int\limits_{ }^{ } \frac{1}{x^2+2}dx\]\[\frac{1}{2}\ln \left| x^2+2 \right|-\int\limits_{ }^{ } \frac{1}{x^2+2}dx\]

Answer 19

looks good! for the other integral you may use table of integrals or try below trig substitution :

\[x = \sqrt{2}\tan (\theta)\]

and recall the trig identity
\[\tan^2\theta + 1 = \sec^2\theta \]

Answer 20

and 1/sqrt{2} d(theta) = dx ?

Answer 21

\[\frac{1}{2}\ln \left| x^2+2 \right|-\frac{1}{2\sqrt{2}}\int\limits_{ }^{ } \frac{1}{ \tan \theta+1}d \theta\]

Answer 22

I took out the 1/2 in the inegral

Answer 23

wait, it is
\[\frac{1}{2}\ln \left| x^2+2 \right|-\frac{1}{2\sqrt{2}}\int\limits_{ }^{ } \frac{1}{ \tan^2\theta+1}d \theta\]

Answer 24

Hmm, I suggest we use table of integrals..

Answer 25

\[\frac{1}{2}\ln \left| x^2+2 \right|-\frac{1}{2\sqrt{2}}\tan^{-1}(\tan \theta +1)+C\]

Answer 26

and then I do substitution

Answer 27

or something wrong?

Answer 28

\[\large \int \dfrac{1}{x^2+a^2}dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)\]

Answer 29

omg, it's 10:30 I haven't even noticed..... lol I got to go. I am really sorry

Answer 30

Hey we are done : \[\int\limits \frac{1}{x^2+2}dx = \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{x}{\sqrt2}\right)\]

Answer 31

we could have worked it easily using the earlier trig substitution but above formula is also good to memorize ^