Solving an Indefinite Integral: \frac{3}{x^2-4}dx

[theakdad]

I have to calculate the following indefinite integral. i know that i have to do it with substitution,but here's my problem,im not enough good to do it.
So integral is :

$$\int \frac{3}{x^2-4}dx$$

 

[MarkFL]

I would consider using partial fraction decomposition, and then the integration is straightforward.

 

[theakdad]

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]

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:

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

Next, assume it may be decomposed as follows:

$$\frac{3}{x^2-4}=\frac{A}{x+2}+\frac{B}{x-2}$$

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

$$3=A(x-2)+B(x+2)$$

Arrange as follows:

$$0x+3=(A+B)x+(2B-2A)$$

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

 

[theakdad]

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:

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

Next, assume it may be decomposed as follows:

$$\frac{3}{x^2-4}=\frac{A}{x+2}+\frac{B}{x-2}$$

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

$$3=A(x-2)+B(x+2)$$

Arrange as follows:

$$0x+3=(A+B)x+(2B-2A)$$

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]

What is $A$ and $B$ representing?

They represent constants, such that:

$$\frac{3}{x^2-4}=\frac{A}{x+2}+\frac{B}{x-2}$$

 

[theakdad]

They represent constants, such that:

$$\frac{3}{x^2-4}=\frac{A}{x+2}+\frac{B}{x-2}$$

Thanks for that,but as we have done it with substitution,i believe i have to do it so.

 

[MarkFL]

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?

 

[theakdad]

Okay, what sort of substitution do you think is appropriate?

lets say $$u=x^2-4$$

 

[MarkFL]

lets say $$u=x^2-4$$

Using this substitution, can you get the correct differential?

 

[theakdad]

Re: Integral calculationdz

Using this substitution, can you get the correct differential?

$$du=2x dz$$ ?

 

[MarkFL]

Re: Integral calculationdz

$$du=2x dz$$ ?

I assume you mean:

$$du=2x\,dx$$

Can you write the original integral as:

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

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...

 

[theakdad]

Re: Integral calculationdz

I assume you mean:

$$du=2x\,dx$$

Can you write the original integral as:

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

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]

Re: Integral calculationdz

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

I suggest:

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

 

[Petrus]

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:

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

Next, assume it may be decomposed as follows:

$$\frac{3}{x^2-4}=\frac{A}{x+2}+\frac{B}{x-2}$$

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

$$3=A(x-2)+B(x+2)$$

Arrange as follows:

$$0x+3=(A+B)x+(2B-2A)$$

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

Hello,
basically 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)$$A+B=0$$
(2)$$2B-2A=3$$
From the (1) we get that $$A=-B$$ put that in (2) and solve for B and Then solve for A cause you know $$A=-B$$
Regards,
$$|\pi\rangle$$

 

[MarkFL]

Hello,
basically 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)$$A+B=0$$
(2)$$2B-2A=3$$
From the (1) we get that $$A=-B$$ put that in (2) and solve for B and Then solve for A cause you know $$A=-B$$
Regards,
$$|\pi\rangle$$

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

 

[theakdad]

Hello,
basically 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)$$A+B=0$$
(2)$$2B-2A=3$$
From the (1) we get that $$A=-B$$ put that in (2) and solve for B and Then solve for A cause you know $$A=-B$$
Regards,
$$|\pi\rangle$$

im stuck with it...its for my homework,that i have to give it today,but I am not into integrals very good,so there will be not much points here...
I should take a deep look into books...

 

[Petrus]

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 a lot early! Well I actually never done trigonometric substitution on this type! Thanks I learned something NEW!:)

Regards,
$$|\pi\rangle$$

 

[Petrus]

im stuck with it...its for my homework,that i have to give it today,but I am 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 interested 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,
$$|\pi\rangle$$

 

[theakdad]

I suggest you do that subsitute as you have not learned the other, if you are interested 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,
$$|\pi\rangle$$

I will take a look! Thank you friend!

 

[MarkFL]

I have to calculate the following indefinite integral. i know that i have to do it with substitution,but here's my problem,im not enough good to do it.
So integral is :

$$\int \frac{3}{x^2-4}dx$$

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

i) Partial fractions:

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

ii) Trigonometric substitution:

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

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

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

And so we obtain:

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

Back substituting for $\theta$, we obtain:

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