T t_reason New member Joined Sep 7, 2006 Messages 2 Sep 7, 2006 #1 Could you please show me how to solve the integral of [(2/(1+x^2)) - |x|] ? I think I need to use u substitution but I'm not sure..
Could you please show me how to solve the integral of [(2/(1+x^2)) - |x|] ? I think I need to use u substitution but I'm not sure..
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Sep 7, 2006 #2 \(\displaystyle \L\\2\int\frac{1}{1+x^{2}}dx+\int{|x|}dx\) Hint: \(\displaystyle \L\\\frac{d}{dx}[tan^{-1}(x)]=\frac{1}{1+x^{2}}\) \(\displaystyle \L\\\int{x}dx=\frac{x^{2}}{2}=\frac{(x)(x)}{2}\;\ and\;\ \int{-x}dx=\frac{-x^{2}}{2}=\frac{(-x)(x)}{2}=\) Therefore, \(\displaystyle \L\\\int{|x|}dx=\frac{x|x|}{2}\)
\(\displaystyle \L\\2\int\frac{1}{1+x^{2}}dx+\int{|x|}dx\) Hint: \(\displaystyle \L\\\frac{d}{dx}[tan^{-1}(x)]=\frac{1}{1+x^{2}}\) \(\displaystyle \L\\\int{x}dx=\frac{x^{2}}{2}=\frac{(x)(x)}{2}\;\ and\;\ \int{-x}dx=\frac{-x^{2}}{2}=\frac{(-x)(x)}{2}=\) Therefore, \(\displaystyle \L\\\int{|x|}dx=\frac{x|x|}{2}\)
T t_reason New member Joined Sep 7, 2006 Messages 2 Sep 7, 2006 #3 thanks so much.. i figured it out! :]
