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

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Answer 1 · 2017-05-01T17:43:32

$"A"(x)=1/3x^3-x^2+4x+"c"$

$intx^2-2x+4$ $dx$

In order to evaluate this integral, apply the reverse power rule to each term separately.

Reverse power rule:

$intx^n$ $dx=1/(n+1)x^(n+1)+ "constant"$

$intx^2-2x+4$ $dx=1/3x^3-2/2x^2+4/1x+"c"$
$=1/3x^3-x^2+4x+"c"$

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