We can split the integral up, as #int(f(x)+-g(x))dx=intf(x)dx+-intg(x)dx:#
#int(12/x^4+8/x^5)dx=int(12/x^4)dx+int(8/x^5)dx#
Let's factor the constants #12# and #8# out of their respective integrals:
#int(12/x^4)dx+int(8/x^5)dx=12intdx/x^4+8intdx/x^5#
Let's rewrite our integrals using negative exponents. Recall that #1/x^a=x^-a.# In other words, bringing a positive exponent from the denominator into the numerator yields the negative form of that exponent.
#12intdx/x^4+8intdx/x^5=12intx^-4dx+8intx^-5dx#
Now we can integrate each of these. Recall that #intx^adx=x^(a+1)/(a+1)+C# where #C# is the constant of integration.
#12intx^-4dx+8intx^-5dx=12(x^(-4+1)/(-4+1))+8(x^(-5+1)/(-5+1))+C=(12x^-3)/-3+(8x^-4)/-4+C=-4x^-3-2x^-2+C#
It may appear confusing that we only have one constant of integration #C# despite taking two integrals. Yes, each of these integrals did originally produce its own distinct constant of integration; however, we combined them both into a single constant #C.# Adding two constants yields one constant; therefore, we just have one constant at the end.
We can revert to positive exponents:
#-4x^-3-2x^-2+C=-4/x^3-2/x^2+C#
Thus,
#int(12/x^4+8/x^5)dx=-4/x^3-2/x^2+C#
