How do you integrate #x(x^2+1)^2 dx#?

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Answer 1 · 2018-03-16T03:15:56

#intx(x^2+1)^2dx=x^6/6+x^4/2+x^2/2+"C"#

Expand #(x^2+1)^2#

#(x^2+1)^2=x^4+2x^2+1#

Distribute the #x#

#x(x^4+2x^2+1)=x^5+2x^3+x#

Next we integrate each term

#intx^5+2x^3+xdx=intx^5dx+int2x^3dx+intxdx#

#=x^6/6+2*x^4/4+x^2/2#

#=x^6/6+x^4/2+x^2/2+"C"#