How to integrate with steps:(4x2-6)/(x+5)(x-2)(3x-1)

[tex]\displaystyle\int\frac{4x^2-6}{(x+5)(x-2)(3x-1)}\,\mathrm dx[/tex]You have a rational expression whose numerator's degree is smaller than the denominator's. This tells you you should consider a partial fraction decomposition. We want to rewrite the integrand in the form[tex]\dfrac{4x^2-6}{(x+5)(x-2)(3x-1)}=\dfrac a{x+5}+\dfrac b{x-2}+\dfrac c{3x-1}[/tex][tex]\implies4x^2-6=a(x-2)(3x-1)+b(x+5)(3x-1)+c(x+5)(x-2)[/tex]You can use the "cover-up" method here to easily solve for [tex]a,b,c[/tex]. It involves fixing a value of [tex]x[/tex] to make 2 of the 3 terms on the right side disappear and leaving a simple algebraic equation to solve for the remaining one.If [tex]x=-5[/tex], then [tex]94=112a\implies a=\dfrac{47}{56}[/tex]If [tex]x=2[/tex], then [tex]10=35b\implies b=\dfrac27[/tex]If [tex]x=\dfrac13[/tex], then [tex]-\dfrac{50}9=-\dfrac{80}9c\implies c=\dfrac58[/tex]So the integral we want to compute is the same as[tex]\displaystyle\frac{47}{56}\int\frac{\mathrm dx}{x+5}+\frac{10}{35}\int\frac{\mathrm dx}{x-2}+\frac58\int\frac{\mathrm dx}{3x-1}[/tex]and each integral here is trivial. We end up with[tex]\displaystyle\int\frac{4x^2-6}{(x+5)(x-2)(3x-1)}\,\mathrm dx=\frac{47}{56}\ln|x+5|+\frac27\ln|x-2|+\frac5{24}\ln|3x-1|+C[/tex]which can be condensed as[tex]\ln\left|(x+5)^{47/56}(x-2)^{2/7}(3x-1)^{5/24}\right|+C[/tex]