How do you evaluate the indefinite integral #int (4x^5-6x^3+7x^2-8)dx#?

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Answer 1 · 2017-10-22T07:02:30

#(2x^6)/3# - #(3x^4)/2# + #(7x^3)/3# - #8x#

#int##4x^5 dx# = #(4x^6)/6 # = #(2x^6)/3#

#int-6x^3 dx# = #(-6x^4)/4# = #(-3x^4)/2#

#int7x^2 dx# = #7x^3/3#

#int-8 dx# = #-8x#

The above answer uses the formula:

#intx^ndx# = #x^(n+1)/(n+1)#

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