Question

a)∫(x^3 −4x^2 −3x +1)dx b)∫(3x^4 −5x^3 +4x +9)dx

Answer 1

$\displaystyle\\a)\int x^3-4x^2-3x+1\ dx=\frac{x^4}{4}-\frac{4x^3}{3}-\frac{3x^2}{2}+x+C\\\\\\ b)\int 3x^4-5x^3+4x +9\ dx=\frac{3x^5}{5}-\frac{5x^4}{4}+4*\frac{x^2}{2}+9x+C=\frac{3x^5}{5}-\frac{5x^4}{4}+\\\\\\ +2x^2+9x+C$