How do you integrate #int (x^3-4x+2)dx#?

1 Answer
Ratnaker Mehta
Feb 5, 2017

#x^4/4-2x^2+2x+C.#

Explanation:

Recall that #intx^ndx=x^(n+1)/(n+1)+c_1, n!=-1,#and, for a constant #k#,

#int[kf(x)+-g(x)]dx=kintf(x)dx+-intg(x)dx+c_2.#

Hence, the reqd. Integral#=intx^3dx-4intx^1dx+2intx^0dx#

#=x^(3+1)/(3+1)-4(x^2/2)+2x#

#=x^4/4-2x^2+2x+C.#

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