Expert: Socrates - 7/31/2006

Question
Integrate (x time sqrt(x^2+3))/(x^2-1)dx.  I can get up to u=x^2-1 and then
1/2 integral sqrt(u+4)/u du.  Can you give me the rest?  

Answer
If you get stuck, try a different substitution.

Let u = (x^2+3)^1/2

du/dx = x(x^2+3)^-1/2

dx = (x^2+3)^1/2 du/x = u/x du

Substitute into S x(x^2+3)^1/2 / (x^2-1) dx

We get S x(u / (x^2-1)) u/x du =

S u^2 / (x^2-1) du

since u = (x^2+3)^1/2 ,

u^2 = x^2+3

u^2-4 = x^2-1

so the integral now becomes

S u^2 / (u^2-4) du =

S (u^2-4) / (u^2-4) + 4 / (u^2-4) du =

S 1 + 4 / (u^2-4) du  =

u + 4 S 1 / (u^2-4) du

Now ,

S 1 / (u^2-4) du =

(1/4) S (1/(u-2) - (1/(u+2)) du =

(1/4) (ln(u-2) - ln(u+2))

so  

u + 4 S 1 / (u^2-4) du =

u + ln(u-2) - ln(u+2)

Now substitute u = (x^2+3)^1/2 and get

(x^2+3)^1/2 + ln((x^2+3)^1/2 - 2 ) - ln((x^2+3)^1/2 + 2)

This is the integral you are looking for. You can check this answer by taking the derivative.

Note that this answer can be simplified a bit by using a rule for logarithms

(x^2+3)^1/2 + ln(((x^2+3)^1/2 - 2 )/((x^2+3)^1/2 + 2))