Question

Calculati
$\int\limits^{2x}_{x}\frac{2t+3}{t+2} \, dt$

Answer 1

$\int\limits^{2x}_{x} \dfrac{2t+3}{t+2} \, dt =\int\limits^{2x}_{x}\dfrac{t+2+t+2-1}{t+2} \, dt = \int\limits^{2x}_{x} \Big(\dfrac{t+2}{t+2}+ \dfrac{t+2}{t+2} - \dfrac{1}{t+2} \Big)\, dt = \\ \\ =\int\limits^{2x}_{x} \Big(1+1-\dfrac{1}{t+2} \Big) \, dt = \int\limits^{2x}_{x}2 \, dt -\int\limits^{2x}_{x} \dfrac{1}{t+2} \, dt = 2t\Big|_x^{2x} -\int\limits^{2x}_{x} \dfrac{(t+2)^'}{t+2} \, dt = \\ \\ =2\cdot 2x-2x - ln(t+2)\Big|_x^{2x} = 4x-2x-ln(2x+2)+ln(x+2) = \\ \\ =2x-ln\Big( \dfrac{2x+2}{x+2}\Big)$