Toán Tìm nguyên hàm của hàm số a) f(x)= (x^2+1)sin2x b) f(x) = x^2+3x-3/x+2 16/09/2021 By Harper Tìm nguyên hàm của hàm số a) f(x)= (x^2+1)sin2x b) f(x) = x^2+3x-3/x+2
Đáp án: $\begin{array}{l}a)I = \int {\left( {{x^2} + 1} \right)\sin 2xdx} \\Dat:\left\{ \begin{array}{l}{x^2} + 1 = u \Rightarrow du = 2xdx\\\sin 2xdx = dv \Rightarrow v = \dfrac{{ – 1}}{2}.cos2x\end{array} \right.\\ \Rightarrow I = \left( {{x^2} + 1} \right).\left( {\dfrac{{ – 1}}{2}.\cos 2x} \right) + \int {\dfrac{1}{2}.\cos 2x.2xdx} \\ = – \dfrac{1}{2}.\left( {{x^2} + 1} \right).cos2x + \int {\cos 2x.x.dx} \\ = – \dfrac{1}{2}.\left( {{x^2} + 1} \right).\cos 2x + x.\dfrac{1}{2}.\sin 2x – \int {\dfrac{1}{2}.\sin 2xdx} \\ = – \dfrac{1}{2}.\left( {{x^2} + 1} \right).\cos 2x + \dfrac{1}{2}.x.\sin 2x + \dfrac{1}{2}.\dfrac{1}{2}.\cos 2x + C\\ = – \dfrac{1}{2}.\left( {{x^2} + 1} \right).\cos 2x + \dfrac{1}{2}.x.\sin 2x + \dfrac{1}{4}.\cos 2x + C\\ = \dfrac{1}{4}.\left( { – 2{x^2} – 1} \right).\cos 2x + \dfrac{1}{2}x.\sin 2x + C\\b)\int {\dfrac{{{x^2} + 3x – 3}}{{x + 2}}dx} \\ = \int {\dfrac{{{x^2} + 4x + 4 – x – 2 – 5}}{{x + 2}}dx} \\ = \int {x + 2 – 1 – \dfrac{5}{{x + 2}}dx} x\\ = \int {x + 1 – \dfrac{5}{{x + 2}}dx} \\ = \dfrac{1}{2}{x^2} + x – 5\ln \left| {x + 2} \right| + C\end{array}$ Trả lời
Đáp án:
$\begin{array}{l}
a)I = \int {\left( {{x^2} + 1} \right)\sin 2xdx} \\
Dat:\left\{ \begin{array}{l}
{x^2} + 1 = u \Rightarrow du = 2xdx\\
\sin 2xdx = dv \Rightarrow v = \dfrac{{ – 1}}{2}.cos2x
\end{array} \right.\\
\Rightarrow I = \left( {{x^2} + 1} \right).\left( {\dfrac{{ – 1}}{2}.\cos 2x} \right) + \int {\dfrac{1}{2}.\cos 2x.2xdx} \\
= – \dfrac{1}{2}.\left( {{x^2} + 1} \right).cos2x + \int {\cos 2x.x.dx} \\
= – \dfrac{1}{2}.\left( {{x^2} + 1} \right).\cos 2x + x.\dfrac{1}{2}.\sin 2x – \int {\dfrac{1}{2}.\sin 2xdx} \\
= – \dfrac{1}{2}.\left( {{x^2} + 1} \right).\cos 2x + \dfrac{1}{2}.x.\sin 2x + \dfrac{1}{2}.\dfrac{1}{2}.\cos 2x + C\\
= – \dfrac{1}{2}.\left( {{x^2} + 1} \right).\cos 2x + \dfrac{1}{2}.x.\sin 2x + \dfrac{1}{4}.\cos 2x + C\\
= \dfrac{1}{4}.\left( { – 2{x^2} – 1} \right).\cos 2x + \dfrac{1}{2}x.\sin 2x + C\\
b)\int {\dfrac{{{x^2} + 3x – 3}}{{x + 2}}dx} \\
= \int {\dfrac{{{x^2} + 4x + 4 – x – 2 – 5}}{{x + 2}}dx} \\
= \int {x + 2 – 1 – \dfrac{5}{{x + 2}}dx} x\\
= \int {x + 1 – \dfrac{5}{{x + 2}}dx} \\
= \dfrac{1}{2}{x^2} + x – 5\ln \left| {x + 2} \right| + C
\end{array}$