What is #int x^4/(x^4-1)dx#? Calculus 1 Answer Tom · Bio Nov 6, 2015 #[x]-1/2[arctan(x)]-1/4[ln|x+1|]+1/4[ln|x-1|]+C# Explanation: #intx^4/(x^4-1)dx = int(x^4-1+1)/(x^4-1)dx = int1dx + int1/(x^4-1)dx# Looking at #int1/(x^4-1)dx# you have #x^4-1=(x^2+1)(x^2-1) = (x^2+1)(x-1)(x+1)# So #int1/((x^2+1)(x+1)(x-1))dx# You need to do partial fraction, you get #-1/2int1/((x^2+1))dx-1/4int1/((x+1))dx+1/4int1/((x-1))dx# #[x]-1/2[arctan(x)]-1/4[ln|x+1|]+1/4[ln|x-1|]+C# Answer link Related questions How do I determine the molecular shape of a molecule? What is the lewis structure for co2? What is the lewis structure for hcn? How is vsepr used to classify molecules? What are the units used for the ideal gas law? How does Charle's law relate to breathing? What is the ideal gas law constant? How do you calculate the ideal gas law constant? How do you find density in the ideal gas law? Does ideal gas law apply to liquids? Impact of this question 863 views around the world You can reuse this answer Creative Commons License