Question

Express the integrand as a sum of partial fractions and evaluate the integral:

Answer 1

\[\int\limits\limits_{?}^{?}\frac{ 8x+x+252 }{x^2+36x }\] no limits, its an indefinite integral

Answer 2

Idk how to put not limits

Answer 3

First you need to factor the denominator. Also, on the top of the fraction do you mean:\[8x^2+x+252?\]

Answer 4

ah yes, sorry, its x^2

Answer 5

For Comparison, im getting \[7\ln |x|-\ln|\frac{ 6 }{\sqrt{x^2+36} }+\frac{ 6 }{\sqrt{x^2+36} }+C\]
Idk if that's right. I tried checking with Wolfram Alpha and Its somewhat different answer

Answer 6

Lets see....\[\frac{8x^2+x+252}{x^2+36x}=\frac{8x^2+x+252}{x(x+36)}=\frac{A}{x}+\frac{B}{x+36}\]This should be the set up for your partial fractions.

Answer 7

Where A and B are some numbers, and we need to figure out what they are.

Answer 8

Im an idiot, apologies, the denominator should read x^3+36x

Answer 9

So that gives me, A=7 B=1 C=1

Answer 10

oh oh oh, then:\[\frac{8x^2+x+252}{x^3+36x}=\frac{8x^2+x+252}{x(x^2+36)}=\frac{A}{x}+\frac{Bx+C}{x^2+36}\] Let me check your A B and C values real fast...

Answer 11

Yeah, that's my setup for the partial fractions

Answer 12

yeah, thats right, your A B and C are right.

Answer 13

okay, so then the first integral would end up being 7ln|x|

Answer 14

So you you have to integrate:\[\frac{7}{x}+\frac{x+1}{x^2+36}.\]That makes the first part of you answer correct.

Answer 15

\[\bf A(x^2+36)+(Bx+C)(x)=8x^2+x+252\]Setting x = 0 we get:\[\bf \bf A(36)=252 \implies A = 7\]Plug this value in for A and re-arrange to get the equation:\[\bf (Bx+C)\cancel{(x)}=x^2+x=\cancel{x}(x+1) \implies Bx + C = x+1 \implies B=C=1\]

Answer 16

yeah, the second integral is what im not sure about. Im using the \[x=a \tan (\theta)\] method

Answer 17

To integrate\[\frac{x+1}{x^2+36}\], you can do:\[\frac{x+1}{x^2+36}=\frac{x}{x^2+36}+\frac{1}{x^2+36}\]The first can be solved by substitution, and the second is:\[\frac{1}{6}\tan^{-1}\left(\frac{x}{6}\right)\]i believe, using the formula:\[\int\limits\frac{1}{x^2+a^2}dx=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C\]

Answer 18

So now the integral becomes:\[\bf 7\int\limits_{}^{}\frac{ 1 }{ x }dx+\int\limits_{}^{}\frac{ x+1 }{ x^2+36 }dx\]

Answer 19

\[\int\limits_{a}^{b} \frac{x+1}{x^2+36}\] = \[\int\limits_{a}^{b} \frac{(6\tan+1)\sec^2(t)}{36\sec^2(t)}\]

Answer 20

by using x=6tan(t) dx=6sec^2(t) and t=tan^-1(x/6)

Answer 21

I see your formula is easier, and yes its probably faster, but they want me to use the other method :(

Answer 22

The first one can be solved using substitution:\[\int\limits\frac{x}{x^2+36}dx=\frac{1}{2}\int\limits\frac{2x}{x^2+36}dx=\frac{1}{2}\int\limits\frac{1}{u}du=\frac{1}{2}\ln\left|u\right|+C\]\[\frac{1}{2}\ln\left|x^2+36\right|+C\]

Answer 23

oh >.< thats the worst

Answer 24

i know! that's why Im confused about it

Answer 25

wats the worst?

Answer 26

So you can cancel out the sec^2(t)'s
@genius when someone forces you to solve a problem a particular way, even though there's a shorter/faster way.

Answer 27

using the trigonometric substitution method. its longer and more painful.

Answer 28

lol that truly sucks...

Answer 29

the sec^2(t) should 6sec^2(t) on the integral i posted before

Answer 30

oh. oops., then it becomes:\[\frac{1}{6}\int\limits6\tan(t)+1dt\]

Answer 31

which is:\[\int\limits\tan(t)+\frac{1}{6}\int\limits1dt\]

Answer 32

the antiderivative of tan is:\[\ln\left|\sec(t)\right|\]

Answer 33

isnt it ln|cos(t)| ?

Answer 34

no its \(\bf ln|sec(t)|\)

Answer 35

which is the same as saying \(\bf -ln|cos(t)|\)

Answer 36

You are missing the negative sign. @Mathard

Answer 37

ohhhh, ty @genius12. That makes sense

Answer 38

So it actually worked out nicely:\[\bf 7\int\limits_{}^{}\frac{ 1 }{ x }dx+\int\limits_{}^{}\frac{ x+1 }{ x^2+36 }dx =7\ln|x|+\ln|\sec(t)|+\frac{ 1 }{ 6 }t+C\]

Answer 39

but we need to get 't' in terms of 'x' now

Answer 40

yeah, we need to substitue back t=tan^-1(x/6)

Answer 41

(Btwy, how do i give medals to @genius12 and @joemath314159 ??) you both are awesome

Answer 42

\[\bf t=\tan^{-1}(x/6)\]Hence:\[\bf =7\ln|x|+\ln|\sec(\tan^{-1}(x/6))|+\frac{ 1 }{ 6 }\tan^{-1}(x/6)+C\]

Answer 43

and, ln|sec(tan^-1(x/6)| = ln|sqrt(x^2+36)/6| no?

Answer 44

\[\ln|\frac{\sqrt{x^2+36}}{6}|\] that looks better

Answer 45

yup @Mathard

Answer 46

okay, so my answer was close. well thats good enough. Thank you so much!

Answer 47

yw

Answer 48

And @joemath314159 was a great help too =]

Answer 49

yes, @joemath314159 thanks a lot :)