Question

Not getting this...

\(I = \int \frac{1}{x^4 - 1} dx\)

Answer 1

@vishweshshrimali5

i know that x^4-1 = (x^2+1) (x+1)(x-1)

Answer 2

but, when we get to this :

\[\large{I = \int \cfrac{1}{(x+1)(x-1)(x^2+1)} }\]

Answer 3

I'd try partial fractions.

Answer 4

\[ I =\cfrac{1}{2} \int \left[ \cfrac{1}{x-1} - \cfrac{1}{x+1} \right] \cfrac{1}{x^2+1} dx \]

Answer 5

No convert all of them into fractions like this:

\[\large{\cfrac{1}{(x+1)(x-1)(x^2+1)} = \cfrac{A}{x+1} + \cfrac{B}{x-1} + \cfrac{Cx + D}{x^2+1}}\]

Answer 6

This is really confusing :

\((A+B+C)x^3 + (B-A+D)x^2 + (A+B-C) x + (B-A-D) = 1\)

Answer 7

this has become more like an cubic equation.

I want all the other terms except (B-A-D) to be zero, right?

Can x be zero?

Answer 8

We are not putting values of x here. It is some general equation. So, try equating the powers of x both sides.

Answer 9

The coefficient of x^3 in LHS was A+B+C but in RHS it is 0.

So, A+B+C = 0.

Answer 10

And so on...

Answer 11

A + B + C = 0

B - A + D = 0

A + B - C = 0

Answer 12

And B- A - D = 1

Answer 13

Yes

Answer 14

Now, that is confusing.

C = 0
A = -B
2B + D = 0
B = -D/2

B = A + D + 1
2B = D + 1
-D/2 = D + 1
3D/2 = -1

D = -2/3

Answer 15

Since,

A + B + C = 0

and A + B - C = 0

Clearly, A + B = C from 2nd

So, 2C = 0 , and C = 0.

Answer 16

A + B = 0 => A = -B ---------(1)

B - A + D = 0

2B + D = 0

B = - D/2

Answer 17

B = A + D + 1

Oh.. okay, got the mistake there.

2B = D + 1

-D = D + 1

D = -1/2

Answer 18

C = 0 ; B = 1/4 ; A = -1/4 ; D = -1/2 ...

Answer 19

Perfect

Answer 20

Yes now you can do simple integration.

Answer 21

I'd go along with this:\[\large{\cfrac{1}{(x+1)(x-1)(x^2+1)} = \cfrac{A}{x+1} + \cfrac{B}{x-1} + \cfrac{Cx + D}{x^2+1}}\] ... Looks like you're already caught up with the task of determining A, B, C and D.

Answer 22

Okay, so that becomes :

\(\cfrac{-1}{4(x+1)} + \cfrac{1}{4(x-1)} - \cfrac{1}{2(x^2 + 1)} \)

Answer 23

Yes, assuming that your A, B, C and D are correct.

Integrate term by term. In your shoes I'd write the first term as \[\frac{ -1 }{ 4 }\int\limits_{a}^{b}\frac{ dx }{ x+1 }.\]

Answer 24

Or leave this integral indefinite and skip the limits of integration.

Answer 25

So, now I have to use this :

\[ \int \cfrac{1}{ax+ b } = \cfrac{1}{a} \log |ax+b| \]

Right

Answer 26

Right

Answer 27

You CAN use that formula, but you don't HAVE TO.

Seeing the following:\[\frac{ -1 }{ 4 }\int\limits\limits_{a}^{b}\frac{ dx }{ x+1 }\]

Answer 28

I would immediately write \[\frac{ -1 }{ 4 }\ln |x+1|+C\] ... but either approach should give you the correct integral.

Answer 29

Oh, yes! Got it sir. That approach will be easier, I guess.

Answer 30

Yeah I would suggest you to do the same as @mathmale sir has suggested to do.

Answer 31

\[\frac{ -1 }{ 4 }\ln |x+1|\]can be re-written as\[\ln |x+1|^{\frac{ -1 }{ 4 }}\] if desired, and the result either left as is or the expression manipulated so that you'd have a positive (instead of negative) exponent. There are other ways of doing this, as well.

Answer 32

Okay, so, I will use this then :

\(\ln (a) - \ln(b) = \ln \left( \cfrac{a}{b} \right) \)

Answer 33

Wouldn't you want to use the law of logarithms that pertains to powers? The rule you've just typed in here pertains to quotients.

Answer 34

It just might help to ignore the coefficients for now and determine, using only rules of logs, what the end result will look like.

THEN, consider the coefficients and determine how they modify your previous result.