Question

Find the indefinite integral.
dy/(y^4+4y^2+4)
Can someone help me?

Answer 1

Can you factor the square?

Answer 2

for the botom

Answer 3

then do you have any ideas what I might suggest next?

Answer 4

\(y^4+4y^2+4=(y^2+2)^2\) so we have $$\int\frac{dy}{(y^2+2)^2}=\dots$$

Answer 5

are you sure the numerator is just \(dy\)?

Answer 6

yeah i'm sure

Answer 7

ok, so now

Answer 8

even though odrin did the factoring for you....(yes i'm looking at you)

Can you think of a method where you split a denominator into two fractions?

Answer 9

Splitting the denominator in 2 will not be of much use here:$$\frac1{(y^2+2)^2}=\frac{A}{y^2+2}+\frac{B}{(y^2+2)^2}\implies A=0,B=1$$

Answer 10

You can salvage the partial fraction decomposition approach by agreeing to factor \(y^2+2=(y-i\sqrt2)(y+i\sqrt2)\) but you'll need to do a decent amount of simplification at the end to get 'nice' results free of any obvious imaginary numbers

Answer 11

...Please stop answering questions I am posing to get the asker thinking. It is very rude

Answer 12

Your suggestion was incorrect.

Answer 13

AGAIN, to get them thinking

Answer 14

you mean split the (y^2+2)^2 into two fractions?

Answer 15

Or, after converting it into (y^2 + 2)^2 , you may simply substitute y = sqrt2 * tan u , to make things easier.

Answer 16

sort of, we have \[\frac{1}{(y^2+2)^2}\] we want to use something called partial fractions, are you familiar with that?

Answer 17

and yea, you could, use that, but why the square root of 2?

Answer 18

ah nvm

Answer 19

I see, for the factoring

Answer 20

i think i'm new with it (partial fractions) ><

Answer 21

@shubhamsrg's suggestion was the one you were looking for @FibonacciChick666 in place of partial fraction decomposition

Answer 22

ok, so if you just covered that, it is probably what they want you to use.

Answer 23

So, odrin's decomp was incorrect

Answer 24

can you fix it?

Answer 25

This may help you look at it @dangbeau https://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/partialfracdirectory/PartialFrac.html

Answer 26

@dangbeau In general, when you have \(y^2+a^2\) without a radical in your denominator, an appropriate trigonometric substitution like \(y=a\tan t\) is your best bet. We know this from the Pythagorean identity:$$\cos^2t+\sin^2t=1\\1+\frac{\sin^2t}{\cos^2 t}=\frac1{\cos^2 t}\\1+\tan^2 t=\sec^2 t$$

Answer 27

Additionally, you may find it convenient to use a hyperbolic trigonometric substitution, as$$\cosh^2t-\sinh^2t=1\\\cosh^2t=1+\sinh^2 t$$i.e. consider \(y=\sqrt2\,\sinh t\), which yields \(dy=\sqrt2\,\cosh t\ dt\):$$\int\frac1{(y^2+2)^2}dy=\int\frac{\sqrt2\,\cosh t}{(2\sinh^2 t+2)^2}dt=\int\frac{\sqrt2\,\cosh t}{4\cosh^4 t}dt=\frac1{\sqrt8}\int\operatorname{sech}^3 t\ dt$$

Answer 28

If you are not familiar with hyperbolic substitutions, this problem will likely not be any easier for you to solve, but if you're savvy with them the above might be a path of less resistance.

Answer 29

@dangbeau , which approach do you like?

Answer 30

hmm this is mutiple choices, and all 4 answers have arctan (y/sqrt2), can u suggest the best way for me to have that answer?

Answer 31

that would be @shubhamsrg 's solution

Answer 32

so long as you have covered the method of that sort of substitution, I think that would be the easiest approach. If not, I'll think on that

Answer 33

i'll try

Answer 34

ok, so list your steps here and we can go over and check them.

Answer 35

@FibonacciChick666 pointed out there is a misleading initial step in my partial fraction decomposition earlier; what I intended was:$$\frac1{(y^2+2)^2}=\frac{Ax+B}{y^2+2}+\frac{Cx+D}{(y^2+2)^2}$$still, however, this invariably yields \(A=B=C=0,D=1\), which does not contradict the earlier post.

Answer 36

\[y=\sqrt{2}tanu \]\[dy=\sqrt{2}\sec ^{2}udu\]=>\[\frac{ \sqrt{2}\sec ^{2}u }{((\sqrt{2}tanu) ^{2}+2)^{2} }\]

Answer 37

good, continue

Answer 38

\[\frac{ \sqrt{2}\tan^{2}u }{ (2\tan ^{2}u +2)^{2} }\]
i'm stucking here T_T

Answer 39

why did you change the top of the fraction to a tangent?

Answer 40

also, should the derivative be on the top or bottom of the fraction?

Answer 41

how to change to a tangent?

Answer 42

? Sorry can you rephrase, I don't understand your question

Answer 43

ah sorry i misread your sentence, i thought you told me change the top to a tangent
so it's \[\frac{ \sqrt{2}\sec ^{2u}du }{(2\tan^{2}u+2)^{2}}\] what should i do next?

Answer 44

so try and factor out those twos if you can

Answer 45

\[\frac{ \sqrt{2}(1+\tan^{2}u )du}{ 4(\tan^{2}u+1)^{2} }\]

Answer 46

now what can you do?

Answer 47

i try thinking but still stuck ><

Answer 48

what is the same?

Answer 49

how to integrate it \[\frac{ 1 }{ \tan^{2}u+1 }du\]

Answer 50

what happened to the sqrt2 and 4?

Answer 51

is it put outside the integral?
\[\frac{ \sqrt{2} }{ 4 }\int\limits_{}^{}\frac{ 1}{ \tan^{2}u+1 }du\]

Answer 52

yea, and now you can make another sub

Answer 53

can you make it more clearly?