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#### wishmaster

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- Oct 11, 2013

- 211

So integral is :

\(\displaystyle \int \frac{3}{x^2-4}dx\)

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- Oct 11, 2013

- 211

So integral is :

\(\displaystyle \int \frac{3}{x^2-4}dx\)

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- #3

- Oct 11, 2013

- 211

it would be good if i would know how to do it....I would consider using partial fraction decomposition, and then the integration is straightforward.

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While there is a shortcut (the Heaviside cover-up method), I recommend you use the following since you are new to it:it would be good if i would know how to do it....

First, factor the denominator:

\(\displaystyle \frac{3}{x^2-4}=\frac{3}{(x+2)(x-2)}\)

Next, assume it may be decomposed as follows:

\(\displaystyle \frac{3}{x^2-4}=\frac{A}{x+2}+\frac{B}{x-2}\)

Then, multiply through by the denominator on the left to get:

\(\displaystyle 3=A(x-2)+B(x+2)\)

Arrange as follows:

\(\displaystyle 0x+3=(A+B)x+(2B-2A)\)

Finally, equate corresponding coefficients to get a 2X2 linear system in $A$ and $B$ which you can then solve.

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- #5

- Oct 11, 2013

- 211

What is $A$ and $B$ representing?While there is a shortcut (the Heaviside cover-up method), I recommend you use the following since you are new to it:

First, factor the denominator:

\(\displaystyle \frac{3}{x^2-4}=\frac{3}{(x+2)(x-2)}\)

Next, assume it may be decomposed as follows:

\(\displaystyle \frac{3}{x^2-4}=\frac{A}{x+2}+\frac{B}{x-2}\)

Then, multiply through by the denominator on the left to get:

\(\displaystyle 3=A(x-2)+B(x+2)\)

Arrange as follows:

\(\displaystyle 0x+3=(A+B)x+(2B-2A)\)

Finally, equate corresponding coefficients to get a 2X2 linear system in $A$ and $B$ which you can then solve.

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- #6

They represent constants, such that:What is $A$ and $B$ representing?

\(\displaystyle \frac{3}{x^2-4}=\frac{A}{x+2}+\frac{B}{x-2}\)

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- #7

- Oct 11, 2013

- 211

Thanks for that,but as we have done it with substitution,i believe i have to do it so.They represent constants, such that:

\(\displaystyle \frac{3}{x^2-4}=\frac{A}{x+2}+\frac{B}{x-2}\)

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Okay, what sort of substitution do you think is appropriate?Thanks for that,but as we have done it with substitution,i believe i have to do it so.

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- #9

- Oct 11, 2013

- 211

lets say \(\displaystyle u=x^2-4\)Okay, what sort of substitution do you think is appropriate?

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Using this substitution, can you get the correct differential?lets say \(\displaystyle u=x^2-4\)

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- #11

- Oct 11, 2013

- 211

\(\displaystyle du=2x dz\) ?Using this substitution, can you get the correct differential?

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- #12

I assume you mean:\(\displaystyle du=2x dz\) ?

\(\displaystyle du=2x\,dx\)

Can you write the original integral as:

\(\displaystyle \int f(u)\,du\) ?

If there was an $x$ as a factor in the numerator of the original integrand, you could, but we don't have that. We are going to need another type of substitution...

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- #13

- Oct 11, 2013

- 211

I assume you mean:

\(\displaystyle du=2x\,dx\)

Can you write the original integral as:

\(\displaystyle \int f(u)\,du\) ?

If there was an $x$ as a factor in the numerator of the original integrand, you could, but we don't have that. We are going to need another type of substitution...

Yes,i mean $dx$.

So what kind of substitution do you suggest?

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- #14

I suggest:Yes,i mean $dx$.

So what kind of substitution do you suggest?

\(\displaystyle x=2\sin(\theta)\)

- Feb 21, 2013

- 739

Hello,While there is a shortcut (the Heaviside cover-up method), I recommend you use the following since you are new to it:

First, factor the denominator:

\(\displaystyle \frac{3}{x^2-4}=\frac{3}{(x+2)(x-2)}\)

Next, assume it may be decomposed as follows:

\(\displaystyle \frac{3}{x^2-4}=\frac{A}{x+2}+\frac{B}{x-2}\)

Then, multiply through by the denominator on the left to get:

\(\displaystyle 3=A(x-2)+B(x+2)\)

Arrange as follows:

\(\displaystyle 0x+3=(A+B)x+(2B-2A)\)

Finally, equate corresponding coefficients to get a 2X2 linear system in $A$ and $B$ which you can then solve.

Basicly this problem subsitute Will NOT give you any progress(there MAY be a way with subsitution (but I Dont think so) AND it can take HUGE time to figoure it out, not worth).. As Mark have helped you it's almost done! You got Two equation (linear)

(1)\(\displaystyle A+B=0\)

(2)\(\displaystyle 2B-2A=3\)

From the (1) we get that \(\displaystyle A=-B\) put that in (2) and solve for B and Then solve for A cause you know \(\displaystyle A=-B\)

Regards,

\(\displaystyle |\pi\rangle\)

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- #16

Hello Petrus!Hello,

Basicly this problem subsitute Will NOT give you any progress(there MAY be a way with subsitution (but I Dont think so) AND it can take HUGE time to figoure it out, not worth).. As Mark have helped you it's almost done! You got Two equation (linear)

(1)\(\displaystyle A+B=0\)

(2)\(\displaystyle 2B-2A=3\)

From the (1) we get that \(\displaystyle A=-B\) put that in (2) and solve for B and Then solve for A cause you know \(\displaystyle A=-B\)

Regards,

\(\displaystyle |\pi\rangle\)

I do agree that partial fractions is much quicker here, but the OP may not have been introduced to this yet. I know when I took Calc II, we were not introduced to partial fractions until after the various substitution methods, although we had seen partial fraction decomposition in Pre-Calculus, we did not apply it to integration until after substitutions.

This integral is doable with the trigonometric substitution I suggested.

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- #17

- Oct 11, 2013

- 211

im stuck with it...its for my homework,that i have to give it today,but im not into integrals very good,so there will be not much points here.......Hello,

Basicly this problem subsitute Will NOT give you any progress(there MAY be a way with subsitution (but I Dont think so) AND it can take HUGE time to figoure it out, not worth).. As Mark have helped you it's almost done! You got Two equation (linear)

(1)\(\displaystyle A+B=0\)

(2)\(\displaystyle 2B-2A=3\)

From the (1) we get that \(\displaystyle A=-B\) put that in (2) and solve for B and Then solve for A cause you know \(\displaystyle A=-B\)

Regards,

\(\displaystyle |\pi\rangle\)

I should take a deep look into books.....

- Feb 21, 2013

- 739

Hello,Hello Petrus!

I do agree that partial fractions is much quicker here, but the OP may not have been introduced to this yet. I know when I took Calc II, we were not introduced to partial fractions until after the various substitution methods, although we had seen partial fraction decomposition in Pre-Calculus, we did not apply it to integration until after substitutions.

This integral is doable with the trigonometric substitution I suggested.

Ohh ok i learned it alot early! Well I actually never done trigonometric substitution on this type! Thanks I learned something NEW!

Regards,

\(\displaystyle |\pi\rangle\)

- Feb 21, 2013

- 739

I suggest you do that subsitute as you have not learned the other, if you are intrested here is a explain how it works (it's not hard) Pauls Online Notes : Calculus II - Partial Fractionsim stuck with it...its for my homework,that i have to give it today,but im not into integrals very good,so there will be not much points here.......

I should take a deep look into books.....

ps. My phone is about to die so i want be able to help soon..

Regards,

\(\displaystyle |\pi\rangle\)

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- #20

- Oct 11, 2013

- 211

I will take a look! Thank you friend!I suggest you do that subsitute as you have not learned the other, if you are intrested here is a explain how it works (it's not hard) Pauls Online Notes : Calculus II - Partial Fractions

ps. My phone is about to die so i want be able to help soon..

Regards,

\(\displaystyle |\pi\rangle\)

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- #21

I will now write out a solution using both methods I suggested:

So integral is :

\(\displaystyle \int \frac{3}{x^2-4}dx\)

\(\displaystyle \frac{3}{4}\int \frac{1}{x-2}-\frac{1}{x+2}\,dx=\frac{3}{4}\ln\left|\frac{x-2}{x+2} \right|+C\)

\(\displaystyle x=2\sin(\theta)\,\therefore\,dx=2\cos(\theta)\, d\theta\)

\(\displaystyle -\frac{3}{4}\int\frac{2\cos(\theta)}{1-\sin^2(\theta)}\,d\theta=-\frac{3}{2}\int \sec(\theta)\,d\theta\)

\(\displaystyle \sec(\theta)\frac{\sec(\theta)+\tan(\theta)}{\sec(\theta)+\tan(\theta)}= \frac{1}{\sec(\theta)+\tan(\theta)} \frac{d}{d\theta}\left(\sec(\theta)+ \tan(\theta) \right)\)

And so we obtain:

\(\displaystyle -\frac{3}{2}\ln\left|\sec(\theta)+\tan(\theta) \right|+C\)

Back substituting for $\theta$, we obtain:

\(\displaystyle -\frac{3}{2}\ln\left|\frac{2}{\sqrt{4-x^2}}+\frac{x}{\sqrt{4-x^2}} \right|+C= -\frac{3}{2}\ln\left|\frac{2+x}{\sqrt{4-x^2}} \right|+C= -\frac{3}{2}\ln\left|\sqrt{\frac{x+2}{x-2}} \right|+C= \frac{3}{4}\ln\left|\frac{x-2}{x+2} \right|+C\)