$\displaystyle \int \frac{x(1+x^2)}{1+x^4)}$ I tried to do it by substitution but reached nowhere
Follow Math Help Forum on Facebook and Google+
Hint: $\displaystyle \int \frac{1}{1+x^2} = tan^{-1}(x)$ oops..
Last edited by Anonymous1; Mar 31st 2010 at 11:25 PM.
Another hint: $\displaystyle \int \frac{x+x^3}{1+x^4} = \int \frac{x}{1+x^4} + \int \frac{x^3}{1+x^4}$
Originally Posted by Anonymous1 Hint: $\displaystyle \int \frac{1}{1+x^4} = tan^{-1}(x^2)$ This is NOT correct. But we can observe that: $\displaystyle \frac{d}{dx} \tan^{-1}(x^2) = \frac{2x}{1+(x^2)^2}$ However, if you didn't recognize this, you'd have to do a bit more work with partial fractions, etc.
Last edited by drumist; Mar 31st 2010 at 11:00 PM.
Originally Posted by roshanhero $\displaystyle \int \frac{x(1+x^2)}{1+x^4)}$ I tried to do it by substitution but reached nowhere Use the substitution $\displaystyle u = x^2$. If you need more help, please show your work and say where you get stuck.
View Tag Cloud