Giải:
a) Ta có \(\int {f\left( x \right)dx} = \int {\left( {3{x^2} - 2x + \frac{2}{x}} \right)dx} = 3\int {{x^2}dx} - 2\int {xdx} + 2\int {\frac{{dx}}{x}} \). \( = 3 \cdot \frac{{{x^3}}}{3} - {x^2} + 2\ln \left| x \right| + C = {x^3} - {x^2} + 2\ln \left| x \right| + C\). b) Ta có \(\int {g\left( x \right)dx} = \int {\left( {\sin x - \frac{3}{{{{\cos }^2}x}} + 1} \right)dx} \) \(= \int {\sin xdx} - 3\int {\frac{1}{{{{\cos }^2}x}}dx} + \int {dx} = - \cos x - 3\tan x + x + C\). c) Ta có \(\int {h\left( x \right)dx} = \int {\left[ {{{\left( {3x - 1} \right)}^2} - 2\sqrt x + \sin x - 1} \right]dx}\) \(= \int {{{\left( {3x - 1} \right)}^2}dx} - 2\int {\sqrt x dx} + \int {\sin dx} - \int {dx} \) \( = \frac{1}{3}\int {{{\left( {3x - 1} \right)}^2}d\left( {3x - 1} \right)} - 2 \cdot \frac{2}{3}x\sqrt x + \cos x - x + C \) \(= \frac{1}{9}{\left( {3x - 1} \right)^3} - \frac{4}{3}x\sqrt x + \cos x - x + C\).Các bài khác cùng chuyên mục