Evaluate (\int \frac{dx}{x(x^3+8)})

Let I = (\int \frac{dx}{x(x^3+8)})

=(\frac{1}{3}\int \frac{3x^2 dx}{x^3(x^3+8)})

Put (x^3+8=t \implies 3x^2 dx = dt)

I = (\frac{1}{3}\int \frac{dt}{t-8} )

=(\frac{1}{3}\int \frac{dt}{t(t-8)})

=(\frac{1}{24}\int \left(\frac{1}{t-8}-\frac{1}{t}\right) dt)

=(\frac{1}{24}[\log|t-8|-\log|t|]+C)

I = (\frac{1}{24}\log\left|\frac{t-8}{t}\right|+C)

=(\frac{1}{24}\log\left|\frac{x^3+8-8}{x^3+8}\right|+C)

=(\frac{1}{24}\log\left|\frac{x^3}{x^3+8}\right|+C)