Tính các tích phân sau

a) \(\int\limits_0^{\frac{\pi }{3}} {(1 + {{\tan }^2}x)dx} \);               b) \[\int\limits_1^2 {\frac{{{x^2} + 2}}{{2{x^2}}}dx} \];                 c) \(\int\limits_0^1 {\frac{{2{\rm{d}}x}}{{{x^2} + 4x + 3}}} \).

Trả lời: 7

Ta có $\underset{1}{\overset{3}{\int }}\frac{x+2}{x}dx=\underset{1}{\overset{3}{\int }}\left(1+\frac{2}{x}\right)dx=\underset{1}{\overset{3}{\int }}dx+\underset{1}{\overset{3}{\int }}\frac{2}{x}dx=2+2{\left.\ln \left|x\right|\right|}_1^3=2+2\ln 3.$

Do đó \(a = 2,\,b = 2,\,c = 3 \Rightarrow S = 7.\)

a) Đ, b) S, c) S, d) S

a) \(\int\limits_a^b {kf\left( x \right)dx} = k\int\limits_a^b {f\left( x \right)dx} = \left. {kF\left( x \right)} \right|_a^b = k\left( {F\left( b \right) - F\left( a \right)} \right)\).

b) \(\int\limits_b^a {f\left( x \right)dx} = \left. {F\left( x \right)} \right|_b^a = F\left( a \right) - F\left( b \right)\).

c) \(S = \int\limits_a^b {\left| {f\left( x \right)} \right|dx} \).

d) \(\int\limits_a^b {f\left( {2x + 3} \right)dx} = \frac{1}{2}\int\limits_a^b {f\left( {2x + 3} \right)d\left( {2x + 3} \right)} = \left. {\frac{1}{2}F\left( {2x + 3} \right)} \right|_a^b\).