# Bài 4 trang 113 SGK Giải tích 12

##### Hướng dẫn giải

Phương pháp tích phân từng phần: $$\int\limits_a^b {udv} = \left. {uv} \right|_a^b - \int\limits_a^b {vdu}$$.

a) Đặt $$\left\{ \begin{array}{l}u = x + 1\\dv = \sin xdx\end{array} \right.$$

b) Đặt $$\left\{ \begin{array}{l}u = \ln x\\dv = {x^2}dx\end{array} \right.$$

c) Đặt $$\left\{ \begin{array}{l}u = \ln \left( {1 + x} \right)\\dv = dx\end{array} \right.$$

d) Đặt $$\left\{ \begin{array}{l}u = {x^2} - 2x - 1\\dv = {e^{ - x}}dx\end{array} \right.$$

Lời giải chi tiết

a) Đặt $$\left\{ \begin{array}{l}u = x + 1\\dv = \sin xdx\end{array} \right.$$ $$\Rightarrow \left\{ \begin{array}{l}du = dx\\v = - \cos x\end{array} \right.$$

$$\begin{array}{l}\Rightarrow \int\limits_0^{\frac{\pi }{2}} {\left( {x + 1} \right)\sin xdx} = \left. { - \left( {x + 1} \right)\cos x} \right|_0^{\frac{\pi }{2}} + \int\limits_0^{\frac{\pi }{2}} {\cos xdx} \\= \left. { - \left( {x + 1} \right)\cos x} \right|_0^{\frac{\pi }{2}} + \left. {\sin x} \right|_0^{\frac{\pi }{2}}\\= 1 + 1 = 2\end{array}$$.

b) Đặt $$\left\{ \begin{array}{l}u = \ln x\\dv = {x^2}dx\end{array} \right.$$ $$\Rightarrow \left\{ \begin{array}{l}du = \frac{{dx}}{x}\\v = \frac{{{x^3}}}{3}\end{array} \right.$$

$$\begin{array}{l}\Rightarrow \int\limits_1^e {{x^2}\ln x} dx = \left. {\left( {\ln x.\frac{{{x^3}}}{3}} \right)} \right|_1^e - \frac{1}{3}\int\limits_1^e {{x^2}dx} \\= \left. {\left( {\ln x.\frac{{{x^3}}}{3}} \right)} \right|_1^e - \left. {\frac{{{x^3}}}{9}} \right|_1^e\\= \frac{{{e^3}}}{3} - \left( {\frac{{{e^3}}}{9} - \frac{1}{9}} \right) = \frac{{2{e^3}}}{9} + \frac{1}{9} = \frac{1}{9}\left( {2{e^3} + 1} \right)\end{array}$$

c) Đặt $$\left\{ \begin{array}{l}u = \ln \left( {1 + x} \right)\\dv = dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \frac{{dx}}{{1 + x}}\\v = x\end{array} \right.$$

$$\begin{array}{l}\Rightarrow \int\limits_0^1 {\ln \left( {x + 1} \right)dx} = \left. {\left( {x.\ln \left( {1 + x} \right)} \right)} \right|_0^1 - \int\limits_0^1 {\frac{x}{{x + 1}}dx} \\= \left. {\left( {x.\ln \left( {1 + x} \right)} \right)} \right|_0^1 - \int\limits_0^1 {\frac{{x + 1 - 1}}{{x + 1}}dx} \\= \left. {\left( {x.\ln \left( {1 + x} \right)} \right)} \right|_0^1 - \int\limits_0^1 {\left( {1 - \frac{1}{{x + 1}}} \right)dx} \\= \left. {\left( {x.\ln \left( {1 + x} \right)} \right)} \right|_0^1 - \left. {\left( {x - \ln \left| {x + 1} \right|} \right)} \right|_0^1\\= \ln 2 - \left( {1 - \ln 2} \right) = 2\ln 2 - 1\end{array}$$

d) Đặt $$\left\{ \begin{array}{l}u = {x^2} - 2x + 1\\dv = {e^{ - x}}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = \left( {2x - 2} \right)dx\\v = - {e^{ - x}}\end{array} \right.$$

$$\begin{array}{l}\Rightarrow \int\limits_0^1 {\left( {{x^2} - 2x - 1} \right){e^{ - x}}dx} = \left. { - {e^{ - x}}\left( {{x^2} - 2x - 1} \right)} \right|_0^1 + 2\int\limits_0^1 {\left( {x - 1} \right){e^{ - x}}dx} \\= \left. { - {e^{ - x}}\left( {{x^2} - 2x - 1} \right)} \right|_0^1 + 2{I_1}\\= 2{e^{ - 1}} - 1 + 2{I_1}\end{array}$$

Đặt $$\left\{ \begin{array}{l}u = x - 1\\dv = {e^{ - x}}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dx\\dv = - {e^{ - x}}\end{array} \right.$$.

$$\begin{array}{l}\Rightarrow {I_1} = \left. { - {e^{ - x}}\left( {x - 1} \right)} \right|_0^1 + \int\limits_0^1 {{e^{ - x}}dx} \\= \left. { - {e^{ - x}}\left( {x - 1} \right)} \right|_0^1\left. { - {e^{ - x}}} \right|_0^1\\= - 1 - \left( {{e^{ - 1}} - 1} \right) =- {e^{ - 1}}\end{array}$$.

Vậy $$I = 2{e^{ - 1}} - 1 - 2{e^{ - 1}} = - 1$$.