Thread: Tricky integral question..need a little info..

Solve: (int) x^2 -x +12/(x^3 + 3x)

This is what I did so far...

x3 + 3x = x(x^2 +3)

= A/x + Bx+C/(x^2 +3)
= A(x^2+3) + Bx+C(x) = (A+B)x^2 + C(x) + 3A
A+B)x^2=1
C=-1
3A=12

so A=4, C=-1 and B = -3

Then....

4/x = 4(int)1/x -3(int) x/x^2 + 3 -1(int) 1/(x^2 + 3) = it's this part I get confused at...I already know what to do for 4(int)1/x but getting the 1/x^2+3 to be 1/(x^2+1) (so it can be arctan) is my problem as I don't know how to make it into a arctan equation..The answer is 4ln(1/x) - 3/2ln(x^2+3) - arctan (1/sqrt 3/(sqrt 3))..do not understand how they got the 3/2ln(x^2+3) as it looks like you have use the tangent function!

Originally Posted by Jgirl689

Solve: (int) x^2 -x +12/(x^3 + 3x)

This is what I did so far...

x3 + 3x = x(x^2 +3)

= A/x + Bx+C/(x^2 +3)
= A(x^2+3) + Bx+C(x) = (A+B)x^2 + C(x) + 3A
A+B)x^2=1
C=-1
3A=12

so A=4, C=-1 and B = -3

Then....

4/x = 4(int)1/x -3(int) x/x^2 + 3 -1(int) 1/(x^2 + 3) = it's this part I get confused at...I already know what to do for 4(int)1/x but getting the 1/x^2+3 to be 1/(x^2+1) (so it can be arctan) is my problem as I don't know how to make it into a arctan equation..The answer is 4ln(1/x) - 3/2ln(x^2+3) - arctan (1/sqrt 3/(sqrt 3))..do not understand how they got the 3/2ln(x^2+3) as it looks like you have use the tangent function!

matching coefficients ...





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