A Ashley5 New member Joined Nov 3, 2007 Messages 14 Nov 20, 2007 #1 x^-3(x^2+4)dx I got u=x^2+4 du=2xdx but how do you sub in for x^-3? is it -3ln(u)? I don't get it
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Nov 21, 2007 #2 Hello, Ashley! \(\displaystyle \int x^{-3}(x^2+4)\,dx\) Click to expand... We have: .\(\displaystyle \displaystyle{\int x^{-3}\left(x^2 + 4\right)\,dx \;=\;\int\left(\frac{1}{x} + 4x^{-3}\right)\,dx} \;=\;\ln x - 2x^{-2} + C\)
Hello, Ashley! \(\displaystyle \int x^{-3}(x^2+4)\,dx\) Click to expand... We have: .\(\displaystyle \displaystyle{\int x^{-3}\left(x^2 + 4\right)\,dx \;=\;\int\left(\frac{1}{x} + 4x^{-3}\right)\,dx} \;=\;\ln x - 2x^{-2} + C\)
