JAMB 2014 · UTME · Q42

Evaluate $\displaystyle\int (2x + 3)^{\frac12}\,dx$ .

Worked solution (try it first)

Add one to the power: $\frac12 + 1 = \frac32$ .
Divide by the new power $\frac32$ and by 2, the derivative of $2x + 3$ : $\dfrac{(2x + 3)^{3/2}}{\frac32 \times 2} = \frac13(2x + 3)^{\frac32}$ .
So the integral is $\frac13(2x + 3)^{\frac32} + k$ , option D.