How do you evaluate the indefinite integral $int (x^2-x+5)dx$ ?

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Answer 1 · 2017-01-22T22:27:45

$int(x^2-x+5)dx=x^3/3-x^2/2+5x+C$

Remember that an indefinite integral is an antiderivative, and since the derivative of sums is the sum of derivatives then

$intf(x)+g(x)+h(x)dx=intf(x)dx+intg(x)dx+inth(x)dx$

it is the same for integrals

In this case

$f(x)=x^2$

$g(x)=-x$

$h(x)=5$

Since $intx^ndx=x^(n+1)/(n+1)$

$intf(x)dx=intx^2dx=x^3/3$

and also since $int-xdx=-intxdx$

$intg(x)dx=int(-x)dx=-intxdx=-x^2/2$

and since for a constant $c$ , $intcdx=cx$ then

$inth(x)dx=int5dx=5x$

Then put them together

$int(x^2-x+5)dx=x^3/3-x^2/2+5x+C$

Answer 2 · 2017-01-22T22:29:14

I tried this:

We can break it into three parts and write:
$intx^2dx-intxdx+int5dx=$

we now use the general integration formula as:
$color(red)(intx^ndx=x^(n+1)/(n+1)+c)$

we can write our integral as:
$intx^2dx-intxdx+5intx^0dx=$
and we get:
$=x^3/3-x^2/2+5x+c$

How do you evaluate the indefinite integral int (x^2-x+5)dxint (x^2-x+5)dx?