Evaluate this indefinite integral

[EvilBunny]

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

 

[Mark44]

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.

 

[EvilBunny]

(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

 

[EvilBunny]

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

 

[Mark44]

(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.

 

[Mark44]

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.