How would you integrate x^2/(3x^2 + 3) dx?
J jeng New member Joined Apr 13, 2006 Messages 6 Apr 27, 2006 #1 How would you integrate x^2/(3x^2 + 3) dx?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Apr 27, 2006 #2 Hello, jeng! How would you integrate: \(\displaystyle \L\,\int\frac{x^2}{3x^2\,+\,3}\,dx\) Click to expand... Do some long division first . . . or: \(\displaystyle \L\,\frac{x^2}{3(x^2\,+\,1)}\:=\:\frac{x^2\,+\,1\,-\,1}{3(x^2\,+\,1)}\;=\;\frac{x^2\,+\,1}{3(x^2\,+\,1)}\,-\,\frac{1}{3(x^2\,+\,1)} \;= \;\frac{1}{3}\,-\,\frac{1}{3}\cdot\frac{1}{x^2\,+\,1}\)
Hello, jeng! How would you integrate: \(\displaystyle \L\,\int\frac{x^2}{3x^2\,+\,3}\,dx\) Click to expand... Do some long division first . . . or: \(\displaystyle \L\,\frac{x^2}{3(x^2\,+\,1)}\:=\:\frac{x^2\,+\,1\,-\,1}{3(x^2\,+\,1)}\;=\;\frac{x^2\,+\,1}{3(x^2\,+\,1)}\,-\,\frac{1}{3(x^2\,+\,1)} \;= \;\frac{1}{3}\,-\,\frac{1}{3}\cdot\frac{1}{x^2\,+\,1}\)
