Question

integral help

Answer 1

\[\int\limits_{}^{}\frac{ 5x+1 }{ x^2-x-12 }\]

Answer 2

Ok.

Answer 3

did you factor the denominator

Answer 4

i would start with that

Answer 5

i managed to get this

Answer 6

\[\frac{ 5x+1 }{ (x-\frac{ 1 }{ 2 })^2-\frac{ 25 }{ 2 } }\]

Answer 7

you don't like partial fractions ?

Answer 8

dont complete the square, thats overkill

Answer 9

i cant figue out how to do the decomposition in this case

Answer 10

the integrand denominator factors nicely, so you dont need to use complete the square

Answer 11

i tried with Ax+B/x^2-x-12

Answer 12

my draw button has invisible ink :o

Answer 13

try
(5x+1) / ( (x-4)(x+3) ) = A / (x-4) + B/(x + 3)

Answer 14

its a refresh problem, try switching the tabs...

Answer 15

i facepalmed so hard right now after getting the proper factors

Answer 16

@ganeshie8 can you be more specific, i should just refresh? or switch what tabs

Answer 17

do you use a tabbed browser like chrome/firefox ?

Answer 18

i use firefox

Answer 19

try below sequence of steps :
1) draw a line
2) goto a different tab
3) come back

Answer 20

didnt work , should i keep the draw button open?

Answer 21

yes...

Answer 22

well thanks for trying :)

Answer 23

do you see a line ?

Answer 24

no, usually the invisible line shows up when u submit...
it is only invisible when u draw it..

Answer 25

oh

Answer 26

bolo, did you find A and B ?

Answer 27

you can use wolfram to find A and B , but its kind of a workaround

Answer 28

a=-2 b=3

Answer 29

so the final answer would be in terms of ln iirc

Answer 30

iirc ?

Answer 31

well, i would have to do it out

Answer 32

iirc=if i remember correctly

Answer 33

thanks, such a simple mistake by not checking the proper factors

Answer 34

dont beat yourself up :)

Answer 35

I have an idea :)
Look at the integrand:
\[\dfrac{5x+1}{x^2-x-12}=\dfrac{4x-2+x+3}{x^2-x-12}\]
\[=\dfrac{4x-2}{x^2-x-12}+\dfrac{x+3}{x^2-x-12}\]
the first term is easy to take integral, right? just let u = x^2 -x -12 to get the answer is 2 ln |u| +C
the second term is \(\dfrac{x+3}{(x+3)(x-4)}=\dfrac{1}{x-4}\)
and it is easy to take integral also, right? :)