Question

\[\int \frac{1+x}{1+x^2}dx\] how??

Answer 1

uhmm i think i agree with that...

Answer 2

use partial fraction decmpostion

Answer 3

hmm \[\frac{1+x}{1+x^2} = \frac{Ax + B}{1+ x^2}?\] i dont think thatll work :/

Answer 4

1/(1+x^2) + x/(1+x^2)
easy integration

Answer 5

what did you do? o.O

Answer 6

ohhh

Answer 7

but 1/1+x^2 doesnt seem easy integration :p

Answer 8

Callisto is awesome

Answer 9

use trigo sub...

Answer 10

1/1+x^2 is always tan^-1(x)

Answer 11

can i do this by u-sub?

Answer 12

ALWASY?

Answer 13

its arctgx

Answer 14

the integral of 1/1+x^2 i mean is always tan^-1(x)
and in the second integral u can use u subst.

Answer 15

i mean:
Integral of 1/1+x^2 = arctg(x) +c

Answer 16

but can i use u sub from the start

Answer 17

yes u can do that

Answer 18

how??

Answer 19

lol im not sure but ill try

Answer 20

this is supposed to be answered by u sub i think

Answer 21

i don't think so

Answer 22

no?

Answer 23

let u=1+x^2
du=2x

Answer 24

arctg(x) +1/2Ln(1+x^2) +c

Answer 25

i dont need the answer -_- im looking for the u sub method

Answer 26

du=2xdx
\[\int\limits{(\frac{1}{1+x^2}+\frac{x}{1+x^2}})dx\]first one is tan^-2(x)
second one
\[\frac{1}{2} \int\limits{\frac{du}{u}}\]

Answer 27

lol you still used trig sub :/

Answer 28

or whatever that's called

Answer 29

is it impossible? :/

Answer 30

there is a general wayto integrate the expression:
\[Mx+N/x ^{2}+px+q\]
where it's supoused that denominator has no real roots

Answer 31

ok...that just became complicated o.O

Answer 32

but that's what you whant, no?

Answer 33

nevermind..ill just go with that solution -_-

Answer 34

if you whant i explain you the method, but it will take for a long time i guess

Answer 35

break this down to:
\[\int\limits (1/x^2+1) + (x/x^2+1) dx\]
\[=\tan^{-1}x+\int\limits x/x^2+1dx\]
let u=x^2+1 du=2xdx du/2=dx
\[=\tan^{-1}x+1/2\int\limits1/udu=\tan^{-1}x+\ln\left| u \right|/2+c\]
\[=\tan^{-1}x+(\ln\left| x^2+1 \right|)/2+c\]

Answer 36

\[\int\limits_{}^{} (1+x)dx/(1+x^{2}) = \int\limits_{}^{} dx/(1+x^{2}) + \int\limits_{}^{} xdx/(1+x^{2})\]

for the one integral use u sub

u = 1 + x^(2)
du = 2xdx
du/2 = xd

thus

\[(1/2) \int\limits_{}^{} du/(u) + \int\limits_{}^{} dx/(1+x^{2}) = (\ln|1+x^{2}|/2) + \int\limits_{}^{} dx/(1+x^{2})\]

for the second integral

we use trigonometric substitution

tan^(2)(x) + 1 = sec^(2)(x)

thus

x = tan(theta)
theta = arctan(x)

now we need to write it in terms of theta

so

d(theta) sec^(2)(theta) = dx
do substitution

\[\int\limits_{}^{} d(\theta)\sec^{2}(\theta)/1 + \tan^{2}(\theta) = \int\limits_{}^{} d(\theta)\sec^{2}/\sec^{2}(\theta) = \int\limits_{}^{} d(\theta) = \theta\]

since theta = arctan(x) we sub that into the answer


and we get

(ln|1+x^(2)|/2) + arctan(x) + c