How do you evaluate the indefinite integral int (2x^2-4x+3)dx?

1 Answer
Guillaume L.
Apr 13, 2018

int(2x^2-4x+3)dx=2/3x^3 -2x^2+3x+C, C in RR

Explanation:

int(2x^2-4x+3)dx=int2x^2dx-int4xdx+int3dx
=2intx^2dx-4intxdx+3intdx
Also, int(x^n)dx=(x^(n+1))/(n+1)
So :int(2x^2-4x+3)dx=2*x^3/3-4*x^2/2+3x+C=2/3x^3-2x^2+3x+C, C in RR
\0/ here's our answer !

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