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Integral calculation

wishmaster

Active member
Oct 11, 2013
211
I have to calculate the following indefinite integral. i know that i have to do it with substitution,but heres my problem,im not enough good to do it.
So integral is :

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would consider using partial fraction decomposition, and then the integration is straightforward.
 

wishmaster

Active member
Oct 11, 2013
211
I would consider using partial fraction decomposition, and then the integration is straightforward.
it would be good if i would know how to do it....
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
it would be good if i would know how to do it....
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.
 

wishmaster

Active member
Oct 11, 2013
211
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.
What is $A$ and $B$ representing?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What is $A$ and $B$ representing?
They represent constants, such that:

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

wishmaster

Active member
Oct 11, 2013
211
They represent constants, such that:

\(\displaystyle \frac{3}{x^2-4}=\frac{A}{x+2}+\frac{B}{x-2}\)
Thanks for that,but as we have done it with substitution,i believe i have to do it so.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Thanks for that,but as we have done it with substitution,i believe i have to do it so.
Okay, what sort of substitution do you think is appropriate?
 

wishmaster

Active member
Oct 11, 2013
211

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

wishmaster

Active member
Oct 11, 2013
211
Re: Integral calculationdz

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Integral calculationdz

\(\displaystyle du=2x dz\) ?
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...
 

wishmaster

Active member
Oct 11, 2013
211
Re: Integral calculationdz

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?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Integral calculationdz

Yes,i mean $dx$.
So what kind of substitution do you suggest?
I suggest:

\(\displaystyle x=2\sin(\theta)\)
 

Petrus

Well-known member
Feb 21, 2013
739
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.
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,
\(\displaystyle |\pi\rangle\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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,
\(\displaystyle |\pi\rangle\)
Hello Petrus! :D

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. :D
 

wishmaster

Active member
Oct 11, 2013
211
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,
\(\displaystyle |\pi\rangle\)
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.......
I should take a deep look into books.....
 

Petrus

Well-known member
Feb 21, 2013
739
Hello Petrus! :D

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. :D
Hello,
Ohh ok i learned it alot early! Well I actually never done trigonometric substitution on this type! Thanks I learned something NEW!:)

Regards,
\(\displaystyle |\pi\rangle\)
 

Petrus

Well-known member
Feb 21, 2013
739
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.......
I should take a deep look into books.....
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,
\(\displaystyle |\pi\rangle\)
 

wishmaster

Active member
Oct 11, 2013
211

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I have to calculate the following indefinite integral. i know that i have to do it with substitution,but heres my problem,im not enough good to do it.
So integral is :

\(\displaystyle \int \frac{3}{x^2-4}dx\)
I will now write out a solution using both methods I suggested:

i) Partial fractions:

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

ii) Trigonometric substitution:

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

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

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

And so we obtain:

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

Back substituting for $\theta$, we obtain:

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