Giải bài 4 trang 101 SGK Giải tích 12

LG a

a) \(∫x\ln (1+x)dx\);

Phương pháp 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:

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

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

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

LG b

b) \(\int {({x^2} + 2x - 1){e^x}dx}\)

Lời giải chi tiết:

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

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

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

LG c

c) \(∫x\sin(2x+1)dx\);

Lời giải chi tiết:

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

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

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

LG d

d) \(\int (1-x)\cos xdx\)

Lời giải chi tiết:

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

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

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

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