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

1 Answer
Oct 22, 2017

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

Explanation:

#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)#

To obtain more reliable answers there are several online calculators available.

Hope my answer helped you.
:)