Question

xdx/4 + 2x + x^2

Answer 1

@hba

Answer 2

\[\int\limits_{}^{}\ (\frac{ x}{ 4 }+2x+x^2)dx\]

Answer 3

^ Is that your question ?

Answer 4

no there is only x dx in numerator

Answer 5

other is denominator

Answer 6

Oh ok lemme correct.

\[\int\limits_{}^{}\frac{ dx }{ 4+2x+x^2 }\]

Answer 7

still u are missing x

Answer 8

\[\int\limits\limits_{}^{}\frac{x dx }{ 4+2x+x^2 }\]

Answer 9

yeah

Answer 10

Any upper or lower limits ?

Answer 11

not without substitution,
first put denominator = u
find du=... ?

Answer 12

x^2+2x+4 = (x+2)^2
So, partial fraction may help

Answer 13

oh... missed that

Answer 14

easiest/shortest is to put u= denominator.

Answer 15

But we have to do it without substituon right?

Answer 16

Yeah we can do it without subs.

Answer 17

x=(x+1)-1
1/2 .2 x = 1/2[ 2x+2]-1
then separate the denominator.

Answer 18

still you'll need substitution.

Answer 19

let me clear!

Answer 20

Yeah you will need subs lol

Answer 21

\[\frac{ 1 }{ 2 }\int\limits \frac{ 2x+2-2 }{ 4+2x+x^2}dx\]

Answer 22

wait friends...

Answer 23

derivative of denominator = 2x+2
so separate it like 2x+2/ denom - 2/denom.

Answer 24

\[\frac{ 1 }{ 2 }\int\limits \frac{ 2x+2 }{ 4+2x+x^2}dx - \int\limits \frac{ 1 }{ x+1)^2+3 }dx\]

Answer 25

@sauravshakya

Answer 26

@hba

Answer 27

Substitution

Answer 28

we can do it without sub. believe me

Answer 29

u=x^2+2x+2

Answer 30

du=2x+2dx

Answer 31

@shubhamsrg

Can you do it without sub ?

Answer 32

Maybe not, best would be to let denominator = u since numerator has 1 degree less than denom

Answer 33

\[\frac{ 1 }{ \sqrt{3} }\tan^{-1} \frac{ x+1 }{ \sqrt{3}} +c\]

Answer 34

just tell me how does it comes

Answer 35

?

Answer 36

this is how its done : (by using standard results)
\[\int \dfrac{f'(x)}{f(x)}dx=\ln |f(x)|+c\]
so first integral is just (1/2)ln(x^2+2x+4)+c
for 2nd , you adjust
\[\int \dfrac{dx}{(x+1)^2+(\sqrt 3)^2}\]
use the result \[\dfrac{dx}{x^2+a^2}=\dfrac{1}{a}\tan^{-1}(x/a)\]
so, you get \[\dfrac{1}{\sqrt3}\tan^{-1}((x+1)/\sqrt3)\]
and thats it!

Answer 37

*
\[\int \dfrac{dx}{x^2+a^2}=\dfrac{1}{a}\tan^{-1}(x/a)\]

Answer 38

+c

Answer 39

@hartnn thankxxx:)

Answer 40

@hba u too:)

Answer 41

Welcome :)

Answer 42

i said u we can do it without sub:L)

Answer 43

Haha :P

Answer 44

do u spend whole of ur time here?

Answer 45

PM me.

Answer 46

yeah okay again thanks so much for help tc bye