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

1. ## 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!

2. 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!
$\frac{x^2-x+12}{x(x^2+3)} = \frac{A}{x} + \frac{Bx+C}{x^2+3}
$

$x^2-x+12 = A(x^2+3) + x(Bx+C)$

matching coefficients ...

$A+B = 1$

$C = -1$

$3A = 12$ ... $A = 4$ , $B = -3$

$\int \frac{4}{x} - \frac{3x+1}{x^2+3} \, dx$

$\int \frac{4}{x} - \frac{3x}{x^2+3} - \frac{1}{x^2+3} \, dx$

note that the antiderivative of $\frac{1}{x^2+a^2}$ is $\frac{1}{a}\arctan\left(\frac{x}{a}\right)$ ...

$4\ln|x| - \frac{3}{2}\ln(x^2+3) - \frac{1}{\sqrt{3}}\arctan\left(\frac{x}{\sqrt{3}}\ right) + C$