Evaluate this indefinite integral

In summary, the indefinite integral has been driving the two people involved nuts. They've tried partial fractions, by parts, substitution, and trig substitution, but are still lost. A polynomial over a polynomial is what they are looking for, but they are lost in trying to identify the legs of a right triangle.
  • #1
EvilBunny
39
0
Evaluate this indefinite integreal

S = integral

S 1/(9+x^2)^2

This has been driving me and my friend nuts.

We tried partial fractions only to realize that it brings us back to the same thing because its not a polynomial over a polynomial, we tried by parts and it did not help and we tried substitution then trig substitution and uhhhh now were lost
 
Physics news on Phys.org
  • #2


It is a polynomial over a polynomial -- 1 is a polynomial of degree 0. You might not have used the right partial fraction decomposition, which would be
(Ax + B)/(9 + x^2) + (Cx + D)/(9 + x^2)^2.

On the other hand, this one is ripe for a trig substitution, I believe, with x/3 = tan(theta). Identify the legs of a right triangle with lengths of x (opposite) and 3 (adjacent). That's the direction I would explore first, if you haven't already done so.
 
  • #3


(Ax + B)/(9 + x^2) + (Cx + D)/(9 + x^2)^2

[ (Ax + B)(9+x^2) + (Cx +D ) ] / ( 9 + x^2 ) ^2 = 1/(9+x^2)^2

(Ax + B)(9+x^2) + (Cx +D ) = 1

9Ax + Ax^3 + 9B + B^2 + Cx + D = 1

Ax^3 + Bx^2 + (9A +C )x + 9B + D = 1

A=0 , B=0 , C=0 , D=1

1/(9+x^2) ^ 2 ??

did i do something wrong
 
  • #4


As for trig sub i come down to

S 1/(U² * sqrt(U-9) )

which would mean my hypotenus is sqrt(U)
and i will have one side 3 and the other sqrt( U-9) ? right

I just never dealt with sqrtU as a side until today
 
  • #5


EvilBunny said:
(Ax + B)/(9 + x^2) + (Cx + D)/(9 + x^2)^2

[ (Ax + B)(9+x^2) + (Cx +D ) ] / ( 9 + x^2 ) ^2 = 1/(9+x^2)^2

(Ax + B)(9+x^2) + (Cx +D ) = 1

9Ax + Ax^3 + 9B + B^2 + Cx + D = 1

Ax^3 + Bx^2 + (9A +C )x + 9B + D = 1

A=0 , B=0 , C=0 , D=1

1/(9+x^2) ^ 2 ??

did i do something wrong

Nope, your work is correct but not helpful. Partial fraction decomposition didn't break the rational expression into two separate rational expressions.
 
  • #6


EvilBunny said:
As for trig sub i come down to

S 1/(U² * sqrt(U-9) )

which would mean my hypotenus is sqrt(U)
and i will have one side 3 and the other sqrt( U-9) ? right

I just never dealt with sqrtU as a side until today

Where is your du? When you do a substitution of any kind you need to replace x and dx with u and du.

With the trig substitution I had in mind, you have

tan u = x/3, so sec^2 (u) *du = dx/3, or dx = 3*sec^2(u)

The hypotenuse is not sqrt(u). The relationship is this: sec^2( u ) = (x^2 + 9)/9.

See if these do you some good.
 

What is an indefinite integral?

An indefinite integral is a mathematical concept used in calculus to find the general antiderivative of a function. It is represented by the symbol ∫ and does not have upper or lower limits, unlike a definite integral.

How do you evaluate an indefinite integral?

To evaluate an indefinite integral, you must use integration techniques such as substitution, integration by parts, or trigonometric substitution. These techniques help to simplify the integral and find the antiderivative of the function.

What is the purpose of evaluating an indefinite integral?

The purpose of evaluating an indefinite integral is to find the general form of the antiderivative of a function. This can be useful in finding the area under a curve, calculating the average value of a function, and solving differential equations.

What are some common mistakes when evaluating an indefinite integral?

Some common mistakes when evaluating an indefinite integral include forgetting the constant of integration, applying the wrong integration technique, and making arithmetic errors. It is important to double-check your work and be careful with each step of the evaluation process.

Can an indefinite integral always be evaluated?

No, not all indefinite integrals can be evaluated analytically. Some integrals are known as "non-elementary" and cannot be expressed using standard mathematical functions. In these cases, numerical methods or approximation techniques may be used to find an approximate value for the integral.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
819
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
829
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
613
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
Back
Top