How do you evaluate the indefinite integral $int (x^3+3x+1)dx$ ?

Question

1289 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-10T09:10:52

The answer is $=x^3/3+3/2x^2+x+C$

We apply

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

So,

$int(x^3+3x+1)dx$

$=intx^3dx+int3xdx+int1dx$

$=x^3/3+3/2x^2+x+C$

How do you evaluate the indefinite integral int (x^3+3x+1)dxint (x^3+3x+1)dx?