int x/sqrt(x^4-6x^2+5) dx Use integration tables to find the indefinite integral.

Recall that indefinite integral follows int f(x) dx = F(x) +C where:
f(x) as the integrand function
F(x) as the antiderivative of f(x)
C as the constant of integration
The given integral problem: int x/(x^4-6x^2+5) dx resembles one of the formulas from the integration table. We follow the integral formula for rational function with roots as:
int (dx)/sqrt(ax^2+bx+c) = 1/sqrt(a)ln|2ax+b+2sqrt(a(ax^2+bx+c))| +C .
For easier comparison, we apply u-substitution by letting: u=x^2
then du= 2x dx or (du)/2 =xdx .
Plug-in the values, we get:
int x/(x^4-6x^2+5) dx =int 1/(x^4-6x^2+5)*x dx
                                 =int 1/(u^2-6u+5)*(du)/2
Apply the basic integration property: int c*f(x) dx = c int f(x) dx .
int 1/(u^2-6u+5)*(du)/2 = 1/2int 1/(u^2-6u+5) du
By comparing ax^2+bx+c  with u^2-6u+5 , we determine the corresponding values as: a=1 , b=-6 ,and c=5 .
Applying the aforementioned formula for rational function with roots, we get:
1/2int 1/(u^2-6u+5) du
=1/2 * [1/sqrt(1)ln|2(1)u+(-6)+2sqrt(1(1u^2+(-6)u+5))|] +C
=1/2 * [1/1ln|2u-6+2sqrt(u^2-6u+5)|] +C
=(ln|2u-6+2sqrt(u^2-6u+5)|)/2 +C
Plug-in u = x^2  and u^2=x^4   on   (ln|2u-6+2sqrt(u^2-6u+5)|)/2 +C , we get the indefinite integral as:
int x/(x^4-6x^2+5) dx =(ln|2x^2-6+2sqrt(x^4-6x^2+5)|)/2 +C