Bài 4 trang 101 SGK Giải tích 12

Đề bài

Sử dụng phương pháp tính nguyên hàm từng phần, hãy tính:

a) \(∫xln(1+x)dx\);             b) \(\int {({x^2} + 2x - 1){e^x}dx}\)

c) \(∫xsin(2x+1)dx\);         d) \(\int (1-x)cosxdx\)

Hướng dẫn giải

Sử dụng phương pháp nguyên hàm từng phần:

Đặt  \(\left\{ \begin{array}{l}u = u\left( x \right)\\dv = v'\left( x \right)dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = u'\left( x \right)dx\\v = v\left( x \right)\end{array} \right..\)

Khi đó ta có:  \(\int {f\left( x \right)dx}  = u\left( x \right)v\left( x \right) - \int {u'\left( x \right)v\left( x \right)dx} .\)

Lời giải chi tiết

 \(a)\;\;\int {x\ln \left( {1 + x} \right)dx.} \)

Đặt:  \(\left\{ \begin{array}{l}u = \ln \left( {1 + x} \right)\\dv = xdx\end{array}\right. \Rightarrow \left\{ \begin{array}{l}du = \frac{1}{{x + 1}}dx\\v = \frac{{{x^2}}}{2}\end{array} \right..\)

 \(\begin{array}{l} \Rightarrow \int {x\ln \left( {1 + x} \right)dx = \frac{{{x^2}}}{2}\ln \left( {1 + x} \right) - \int {\frac{{{x^2}}}{{2\left( {x + 1} \right)}}dx} } \\ = \frac{{{x^2}}}{2}\ln \left( {1 + x} \right) - \frac{1}{2}\int {\left( {\frac{{{x^2} - 1}}{{x + 1}} + \frac{1}{{x + 1}}} \right)dx} \\ = \frac{{{x^2}}}{2}\ln \left( {1 + x} \right) - \frac{1}{2}\int {\left( {x - 1 + \frac{1}{{x + 1}}} \right)dx} \\= \frac{{{x^2}}}{2}\ln \left( {1 + x} \right) - \frac{1}{2}\left( {\frac{{{x^2}}}{2} - x + \ln \left( {1 + x} \right)} \right) + C\\ = \frac{{{x^2}}}{2}\ln \left( {1 + x} \right) - \frac{{{x^2}}}{4} + \frac{x}{2} - \frac{1}{2}\ln \left( {1 + x} \right) + C\\= \frac{1}{2}\left( {{x^2} - 1} \right)\ln \left( {1 + x} \right) - \frac{{{x^2}}}{4} + \frac{x}{2} + C.\end{array}\)

 \(b)\;\int {\left( {{x^2} + 2x - 1} \right){e^x}dx.} \)

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

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

Xét  \(\int {\left( {x + 1} \right){e^x}dx:} \)

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

\(\begin{array}{l}\Rightarrow \int {\left( {x + 1} \right){e^x}dx}  = \left( {x + 1} \right){e^x} - \int {{e^x}dx} \\ = \left( {x + 1} \right){e^x} - {e^x} + C = x{e^x} + C.\\ \Rightarrow \int {\left( {{x^2} + 2x - 1} \right){e^x}dx}  = \left( {{x^2} + 2x - 1} \right){e^x} - 2x{e^x} + C\\ = \left( {{x^2} - 1} \right){e^x} + C.\end{array}\)

\(c)\;\;\int {x\sin \left( {2x + 1} \right)dx} .\)

Đặt:  \(\left\{ \begin{array}{l}u = x\\dv = \sin \left( {2x + 1} \right)dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dx\\v =  - \frac{1}{2}\cos \left( {2x + 1} \right)\end{array} \right..\)

\(\begin{array}{l} \Rightarrow \int {x\sin \left( {2x + 1} \right)dx}  =  - \frac{1}{2}x\cos \left( {2x + 1} \right) + \frac{1}{2}\int {\cos \left( {2x + 1} \right)dx} \\ =  - \frac{1}{2}x\cos \left( {2x + 1} \right) + \frac{1}{4}\sin \left( {2x + 1} \right) + C.\end{array}\)

\(d)\;\;\int {\left( {1 - x} \right)\cos xdx} \)

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

\(\begin{array}{l}\Rightarrow \int {\left( {1 - x} \right)\cos xdx}  = \left( {1 - x} \right)\sin x + \int {\sin xdx} \\= \left( {1 - x} \right)\sin x - \cos x + C.\end{array}\)