Question

a little help with one medium/hard integral?

Answer 1

Have you tried partial fractions?

Answer 2

oh? like divide the top by bottom? D:
I was asked to do this with u sub but I dunno how

Answer 3

Not quite. What I mean is split up the given rational function into component rational functions, like you would with
\[\frac{x}{(x-1)(x+1)}=\frac{1}{2(x-1)}+\frac{1}{2(x+1)}\]

Answer 4

you can use u-sub

Answer 5

Yes instructions want me to use u sub :O

Answer 6

Even so, partial fractions will probably be needed. A single substitution won't take care of the integral in one step.

Answer 7

Let's say you want to go the substitution route. You'd try \(u=x^4+x\), which gives \(du=(4x^3+1)~dx\). You'll want to rewrite the original integrand so you can apply this sub:
\[\int\frac{2x^3-1}{x^4+x}~dx=\frac{1}{2}\int\frac{4x^3+1}{x^4+x}~dx-\frac{3}{2}\int\frac{dx}{x^4+x}\]

Answer 8

hm? we're substituting u right into this equation now?

Answer 9

We could, for the first integral. You'll reduce it to a logarithmic result. Partial fractions for the second.

Answer 10

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

Answer 11

ahh ok, I think I get this step :) What can I do after?

Answer 12

Factorizing the denominator of the second integral is the next step:
\[x^4+x=x(x^3+1)=x(x+1)(x^2-x+1)\]

Answer 13

So,
\[\frac{1}{x(x+1)(x^2-x+1)}=\frac{a}{x}+\frac{b}{x+1}+\frac{cx+d}{x^2-x+1}\]
and solve for the constants as you normally would.

Answer 14

try this and see if it'll be easier to use u-sub without integration gymnastics \[\int\limits \frac{ 2x^3 -1 }{ x^4+x }dx \rightarrow \int\limits \frac{2x- \frac{1}{x^2}}{x^2+ \frac{1}{x}} dx\]

Answer 15

I prefer the most tedious route :P

Answer 16

haha I figured. I hope I didn't intrude

Answer 17

wait ninja can you explain your ways too, please?

Answer 18

@sith may I ask what is c and d?

Answer 19

nice trick haha :)
so many stones to hit our heads with :

we can use the similar trick for second integral in S&G's route :
\[\int \dfrac{1}{x^4+x}dx = \int \dfrac{(x^3+1)-x^3}{x(x^3+1)}dx = \int \dfrac{1}{x}dx - \int \dfrac{x^2}{x^3+1}dx \]

Answer 20

sorry nin****

Answer 21

oh then second intergral would be ln(x) -x -arc cot (x) + C

Answer 22

what about the first? I'm a little confused on the cx+d part

Answer 23

it is a pure u-sub
it is just rewriting before we actually use any integration technique

just as
1/x is the same as x^(-1)
we can always rewrite our problem

Answer 24

@nincompoop not at all! Any more efficient method is always welcome.

@study100 \(c\) and \(d\) are coefficients of a linear factor. If you haven't learned about partial fractions, or haven't been exposed to higher order factors in the denominator, you can rely on the others' methods.

If you're curious: \(x^2-x+1\) is an irreducible quadratic factor (no real roots, so you can't write it in \((x-x_1)(x-x_2)\) form. Instead of using a constant factor, we need to use a linear one. It's kind of a "this is how it is" thing to remember for me, so I'm not sure how else to explain it. There are plenty of resources online you can refer to if you'd like to see why we do it.

Answer 25

I somewhat get the definition now that I googled it up a bit while waiting.
Sorry, would you give me a few more hints on the 1st one?

Answer 26

@nincompoop what would be the u sub with your method?

Answer 27

you use the denominator as your u

Answer 28

u= x^2 + 1/x ?
du = 2x -1/x^2

Answer 29

\[u = x^2 + \frac{1}{x} \]

Answer 30

yes, try that and see the magic appear before your eyes

Answer 31

1st integral = 1/2 ln u? for your method @SithsAndGiggles

Answer 32

Yes, \(\dfrac{1}{2}\ln u=\dfrac{1}{2}\ln(x^4+x)\).

Answer 33

holy moly
integral of du/u : o what

Answer 34

laughing out loud

Answer 35

so answer in sith's method
f(x) = 1/2 ln(x^4+x) - 3/2 ( ln(x) -x - arc cot (x) ) + C

Answer 36

and nin's method is ln (u)
which is ln (x^2 + (1/x)) + C

Answer 37

it's different, but maybe the same after simplification later?

Answer 38

should be

Answer 39

and holy moly nin :o how did you change the fractions to that form?

Answer 40

You can use this property to keep rewriting to your heart's content:\[\ln(x^4+x)=\ln [x(x^3+1)]=\ln x+\ln(x^3+1)\]

Answer 41

how?
practice, practice and practice
http://finedrafts.com/files/CUNY/math/calculus/Larson%20etal%209th%20ed/

Answer 42

Wow, I bookmarked that link , ty nin.
I'll try to rewrite this using that property for practice too.

Good god, thank you so much! @SithsAndGiggles @nincompoop

Answer 43

You're welcome, but the medal distribution is a bit skewed :)

Answer 44

Closed :) <3 ty @ganeshie8 for helping with 2nd integral too

Answer 45

there, for nin too :D

Answer 46

in the future, you will be forced to do partial fractions and all sorts of integration gymnastics for even the easiest integration problem
I had to go through it and I bet even the other students here did as well

Answer 47

I guess there's no way out of it v.v
does that site link you gave me includes practice for partial fractions?

Answer 48

Idk, I'm self taught on these topics so I just look at hotmath and online textbooks to learn calc :o

Answer 49

Wonder if someone can recommend a book

Answer 50

same with me :3 but i feel you can learn everything about partial fractions in less than 1 hour :
http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/lecture-29-partial-fractions/

somewhere in the mid of video, the prof talks about `cover up` method, which is really a cool way to find the coefficients... have fun :)

Answer 51

@ganeshie8 thank you so much! I'll start studying with this vid right away >u< <3