How would you find the integral of x/(36+x^4)...had a student ask me and can't figure it out for some reason :\
Follow Math Help Forum on Facebook and Google+
Let $\displaystyle t = x^2 $ to get: $\displaystyle \displaystyle \int \frac{x}{36+(x^2)^2}\;{dx} = \frac{1}{2}\int\frac{1}{6^2+t^2}\;{dt} = \frac{1}{12}\arctan\frac{t}{6}+k = \frac{1}{12}\arctan\frac{x^2}{6}+k$.
View Tag Cloud