Find ∫(x2 + 3x − 5)dx

A.

\(\frac{x_3}{3}\) - \(\frac{3x_2}{2}\) - 5x + k

B.

\(\frac{x_3}{3}\) - \(\frac{3x_2}{2}\) + 5x + k

C.

\(\frac{x_3}{3}\) + \(\frac{3x_2}{2}\) - 5x + k

D.

\(\frac{x_3}{3}\) + \(\frac{3x_2}{2}\) + 5x + k

Correct answer is C

∫xndx = \(\frac{x_{n + 1}}{n + 1}\)

∫dx = x + k

where k is constant

∫(x2 + 3x − 5)dx

∫x2 dx + ∫3xdx − ∫5dx

\(\frac{2_{2 + 1}}{2 + 1}\) + \(\frac{3x^{1 + 1}}{1 + 1}\) − 5x + k

\(\frac{x_3}{3}\) + \(\frac{3x_2}{2}\) − 5x + k