Nguyên hàm \[I = \int {{e^x}\sin xdx} \] là:

Hướng dẫn giải

Phân tích \[f\left( x \right) = \frac{{2x + 1}}{{{x^4} + 2{x^3} + {x^2}}} = \frac{{2x + 1}}{{{x^2}{{\left( {x + 1} \right)}^2}}} = \frac{{2x + 1}}{{{{\left( {{x^2} + x} \right)}^2}}}\]

Khi đó \[F\left( x \right) = \int {\frac{{2x + 1}}{{{{\left( {{x^2} + x} \right)}^2}}}dx} = \int {\frac{1}{{{{\left( {{x^2} + x} \right)}^2}}}d\left( {{x^2} + x} \right)} = - \frac{1}{{{x^2} + x}} + C\].

Mặt khác \[F\left( 1 \right) = \frac{1}{2} \Rightarrow - \frac{1}{2} + C = \frac{1}{2} \Rightarrow C = 1\].

Vậy \[F\left( x \right) = - \frac{1}{{{x^2} + x}} + 1 = - \frac{1}{{x\left( {x + 1} \right)}} + 1 = - \left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right) + 1\].

Do đó \[\begin{array}{l}S = F\left( 1 \right) + F\left( 2 \right) + F\left( 3 \right) + ... + F\left( {2019} \right) = - \left( {1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{2019}} - \frac{1}{{2020}}} \right) + 2019\\\;\;\; = - \left( {1 - \frac{1}{{2020}}} \right) + 2019 = 2018 + \frac{1}{{2020}} = 2018\frac{1}{{2020}}\end{array}\]

Chọn C.

Hướng dẫn giải

\[f\left( x \right) = \int {f'\left( x \right)dx} = \int {\frac{2}{{2x - 1}}dx} = \ln \left| {2x - 1} \right| + C = \left\{ \begin{array}{l}\ln \left( {2x - 1} \right) + {C_1}\;khi\;x > \frac{1}{2}\\\ln \left( {1 - 2x} \right) + {C_2}\;khi\;x < \frac{1}{2}\end{array} \right.\]

Vì \[\left\{ \begin{array}{l}f\left( 0 \right) = 1\\f\left( 1 \right) = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{C_2} = 1\\{C_1} = 2\end{array} \right.\].

Suy ra \[f\left( x \right) = \left\{ \begin{array}{l}\ln \left( {2x - 1} \right) + 2\;khi\;x > \frac{1}{2}\\\ln \left( {1 - 2x} \right) + 1\;khi\;x < \frac{1}{2}\end{array} \right.\].

Do đó  \[P = f\left( { - 1} \right) + f\left( 3 \right) = 3 + \ln 3 + \ln 5 = 3 + \ln 15\]

Chọn D.