giải bài tập sgk giải tích 12 cơ bản: tích phân

Bài viết hướng dẫn giải các bài tập trong phần câu hỏi và bài tập và phần luyện tập của sách giáo khoa Giải tích 12 cơ bản: Tích phân.

CÂU HỎI VÀ BÀI TẬP

Bài 1. Tính các tích phân sau đây:
a) \(\int_{ – \frac{1}{2}}^{\frac{1}{2}} {\sqrt[3]{{{{(1 – x)}^2}}}} dx.\)
b) \(\int_0^{\frac{\pi }{2}} {\sin } \left( {\frac{\pi }{4} – x} \right)dx.\)
c) \(\int_{\frac{1}{2}}^2 {\frac{1}{{x(x + 1)}}dx} .\)
d) \(\int_0^2 x {(x + 1)^2}dx.\)
e) \(\int_{\frac{1}{2}}^2 {\frac{{1 – 3x}}{{{{(x + 1)}^2}}}dx} .\)
f) \(\int_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} {\sin } 3x.\cos 5xdx.\)

Lời giải:
a) Đặt \(u = 1 – x\) ta có \(du = – dx.\)
Khi \(x = – \frac{1}{2}\) thì \(u = \frac{3}{2}\); khi \(x = \frac{1}{2}\) thì \(u = \frac{1}{2}.\)
Do đó: \(\int_{ – \frac{1}{2}}^{\frac{1}{2}} {\sqrt[3]{{{{(1 – x)}^2}}}dx} \) \( = – \int_{\frac{3}{2}}^{\frac{1}{2}} {\sqrt[3]{{{u^2}}}} du\) \( = \int_{\frac{1}{2}}^{\frac{3}{2}} {{u^{\frac{2}{3}}}} du\) \( = \left. {\frac{3}{5}{u^{\frac{5}{3}}}} \right|_{\frac{1}{2}}^{\frac{3}{2}}\) \( = \left. {\frac{3}{5}u\sqrt[3]{{{u^2}}}} \right|_{\frac{1}{2}}^{\frac{3}{2}}.\)
\( = \frac{3}{5}\left( {\frac{3}{2}\sqrt[3]{{\frac{9}{4}}} – \frac{1}{2}\sqrt[3]{{\frac{1}{4}}}} \right)\) \( = \frac{3}{{10\sqrt[3]{4}}}(3\sqrt[3]{9} – 1).\)
b) Đặt \(u = \frac{\pi }{4} – x\) ta có \(du = – dx.\)
Khi \(x = 0\) thì \(u = \frac{\pi }{4}\); khi \(x = \frac{\pi }{2}\) thì \(u = – \frac{\pi }{4}.\)
Do đó: \(\int_0^{\frac{\pi }{2}} {\sin } \left( {\frac{\pi }{4} – x} \right)dx\) \( = – \int_{\frac{\pi }{4}}^{ – \frac{\pi }{4}} {\sin udu} \) \( = \int_{ – \frac{\pi }{4}}^{\frac{\pi }{4}} {\sin udu} \) \( = – \left. {\cos u} \right|_{ – \frac{\pi }{4}}^{\frac{\pi }{4}}\) \( = – \left( {\cos \frac{\pi }{4} – \cos \left( { – \frac{\pi }{4}} \right)} \right) = 0.\)
Vậy \(\int_0^{\frac{\pi }{2}} {\sin } \left( {\frac{\pi }{4} – x} \right)dx = 0.\)
c) Ta có: \(\frac{1}{{x(x + 1)}} = \frac{1}{x} – \frac{1}{{x + 1}}.\)
Do đó: \(\int_{\frac{1}{2}}^2 {\frac{{dx}}{{x(x + 1)}}} \) \( = \int_{\frac{1}{2}}^2 {\left( {\frac{1}{x} – \frac{1}{{x + 1}}} \right)dx} \) \( = \int_{\frac{1}{2}}^2 {\frac{{dx}}{x}} – \int_{\frac{1}{2}}^2 {\frac{{dx}}{{x + 1}}} \) \( = \int_{\frac{1}{2}}^2 {\frac{{dx}}{x}} – \int_{\frac{1}{2}}^2 {\frac{{d(x + 1)}}{{x + 1}}} \) \( = \left. {\ln |x|} \right|_{\frac{1}{2}}^2 – \left. {\ln |x + 1|} \right|_{\frac{1}{2}}^2.\)
\( = \ln 2 – \ln \frac{1}{2} – \ln 3 – \ln \frac{3}{2}\) \( = \ln 2.\)
d) \(\int_0^2 x {(x + 1)^2}dx\) \( = \int_0^2 {\left( {{x^3} + 2{x^2} + x} \right)dx} \) \( = \left. {\left( {\frac{{{x^4}}}{4} + \frac{2}{3}{x^3} + \frac{1}{2}{x^2}} \right)} \right|_0^2\) \( = 4 + \frac{{16}}{3} + 2\) \( = \frac{{34}}{3}.\)
e) Đặt \(u = x + 1\) ta có \(du = dx\) và \(x = u – 1.\)
Khi \(x = \frac{1}{2}\) thì \(u = \frac{3}{2}\); khi \(x = 2\) thì \(u = 3.\)
Do đó: \(\int_{\frac{1}{2}}^2 {\frac{{1 – 3x}}{{{{(x + 1)}^2}}}dx} \) \( = \int_{\frac{3}{2}}^3 {\frac{{1 – 3(u – 1)}}{{{u^2}}}du} \) \( = \int_{\frac{3}{2}}^3 {\frac{{4 – 3u}}{{{u^2}}}du} \) \( = 4\int_{\frac{3}{2}}^3 {\frac{{du}}{{{u^2}}}} – 3\int_{\frac{3}{2}}^3 {\frac{{du}}{u}} \) \( = – \left. {\frac{4}{u}} \right|_{\frac{3}{2}}^3 – \left. {3\ln u} \right|_{\frac{3}{2}}^3.\)
\( = – \left( {\frac{4}{3} – \frac{4}{{\frac{3}{2}}}} \right) – 3\left( {\ln 3 – \ln \frac{3}{2}} \right)\) \( = \frac{4}{3} – 3\ln 2.\)
f) Ta có: \(\sin 3x.\cos 5x\) \( = \frac{1}{2}(\sin 8x – \sin 2x).\)
Do đó: \(\int_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} {\sin } 3x.\cos 5xdx\) \( = \frac{1}{2}\int_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} {(\sin 8x – \sin 2x)dx} \) \( = \frac{1}{2}\int_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} {\sin 8xdx} – \frac{1}{2}\int_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} {\sin 2xdx} .\)
\( = \left. { – \frac{1}{{16}}\cos 8x} \right|_{ – \frac{\pi }{2}}^{\frac{\pi }{2}} + \left. {\frac{1}{4}\cos 2x} \right|_{ – \frac{\pi }{2}}^{\frac{\pi }{2}}\) \( = – \frac{1}{{16}}[\cos 4\pi – \cos ( – 4\pi )]\) \( + \frac{1}{4}[\cos \pi – \cos ( – \pi )].\)
\( = – \frac{1}{{16}}(1 – 1) + \frac{1}{4}( – 1 + 1) = 0.\)

Bài 2. Tính các tích phân sau:
a) \(\int_0^2 {|1 – x|dx} .\)
b) \(\int_0^{\frac{\pi }{2}} {{{\sin }^2}xdx} .\)
c) \(\int_0^{\ln 2} {\frac{{{e^{2x + 1}} + 1}}{{{e^x}}}dx.} \)
d) \(\int_0^\pi {\sin 2x.{{\cos }^2}xdx.} \)

Lời giải:
a) Ta có: \(|1 – x|\) \( = \left\{ {\begin{array}{*{20}{l}}
{1 – x\:{\rm{ với }}\:0 \le x \le 1}\\
{x – 1\:{\rm{ với }}\:1 \le x \le 2}
\end{array}} \right..\)
Do đó:
\(\int_0^2 {|1 – x|dx} \) \( = \int_0^1 {(1 – x)dx} + \int_1^2 {(x – 1)dx} \) \( = \left. {\left( {x – \frac{{{x^2}}}{2}} \right)} \right|_0^1 + \left. {\left( {\frac{{{x^2}}}{2} – x} \right)} \right|_1^2\) \( = \frac{1}{2} + \frac{1}{2} = 1.\)
b) Ta có: \({\sin ^2}x = \frac{{1 – \cos 2x}}{2}.\)
Do đó: \(\int_0^{\frac{\pi }{2}} {{{\sin }^2}xdx} \) \( = \frac{1}{2}\int_0^{\frac{\pi }{2}} {(1 – \cos 2x)dx} \) \( = \frac{1}{2}\int_0^{\frac{\pi }{2}} {dx} – \frac{1}{2}\int_0^{\frac{\pi }{2}} {\cos 2xdx} \) \( = \left. {\frac{1}{2}x} \right|_0^{\frac{\pi }{2}} – \left. {\frac{1}{4}\sin 2x} \right|_0^{\frac{\pi }{2}}\) \( = \frac{\pi }{4}.\)
c) \(\int_0^{\ln 2} {\frac{{{e^{2x + 1}} + 1}}{{{e^x}}}dx} \) \( = \int_0^{\ln 2} {\left( {{e^{x + 1}} + {e^{ – x}}} \right)dx} \) \( = \int_0^{\ln 2} {{e^{x + 1}}} dx + \int_0^{\ln 2} {{e^{ – x}}} dx.\)
\( = \int_0^{\ln 2} {{e^{x + 1}}} d(x + 1) – \int_0^{\ln 2} {{e^{ – x}}} d( – x)\) \( = \left. {{e^{x + 1}}} \right|_0^{\ln 2} – \left. {{e^{ – x}}} \right|_0^{\ln 2}.\)
\( = {e^{\ln 2 + 1}} – {e^1} – {e^{ – \ln 2}} + 1\) \( = {e^{\ln 2 + 1}} – \frac{1}{{{e^{\ln 2}}}} – (e – 1)\) \( = e + \frac{1}{2}.\)
d) \(\int_0^\pi {\sin 2x.{{\cos }^2}xdx} \) \( = 2\int_0^\pi {\sin x.{{\cos }^3}xdx} \) \( = – 2\int_0^\pi {{{\cos }^3}xd(\cos x)} \) \( = – \left. {2.\frac{1}{4}{{\cos }^4}x} \right|_0^\pi .\)
\( = – \frac{1}{2}\left( {{{\cos }^4}\pi – {{\cos }^4}0} \right) = 0.\)

Bài 3. Sử dụng phương pháp đổi biến số, hãy tính:
a) \(\int_0^3 {\frac{{{x^2}}}{{{{(1 + x)}^{\frac{3}{2}}}}}dx} .\)
b) \(\int_0^1 {\sqrt {1 – {x^2}} } dx.\)
c) \(\int_0^1 {\frac{{{e^x}(1 + x)}}{{1 + x{e^x}}}dx.} \)
d) \(\int_0^{\frac{a}{2}} {\frac{1}{{\sqrt {{a^2} – {x^2}} }}dx} \) \((a > 0).\)

Lời giải:
a) Đặt \(u = 1 + x\) ta có \(du = dx\); \({x^2} = {(u – 1)^2}.\)
Khi \(x = 0\) thì \(u = 1\); khi \(x = 3\) thì \(u = 4.\)
Do đó: \(\int_0^3 {\frac{{{x^2}}}{{{{(1 + x)}^{\frac{3}{2}}}}}dx} \) \( = \int_1^4 {\frac{{{{(u – 1)}^2}}}{{{u^{\frac{3}{2}}}}}du} \) \( = \int_1^4 {\frac{{{u^2} – 2u + 1}}{{{u^{\frac{3}{2}}}}}du} .\)
\( = \int_1^4 {\left( {{u^{\frac{1}{2}}} – 2{u^{ – \frac{1}{2}}} + {u^{ – \frac{3}{2}}}} \right)du} \) \( = \int_1^4 {{u^{\frac{1}{2}}}} du – 2\int_1^4 {{u^{ – \frac{1}{2}}}} du + \int_1^4 {{u^{ – \frac{3}{2}}}} du.\)
\( = \left. {\frac{2}{3}{u^{\frac{3}{2}}}} \right|_1^4 – \left. {4{u^{\frac{1}{2}}}} \right|_1^4 – \left. {2{u^{ – \frac{1}{2}}}} \right|_1^4\) \( = \frac{{16}}{3} – \frac{2}{3} – (8 – 4) – 2\left( {\frac{1}{2} – 1} \right)\) \( = \frac{{14}}{3} – 3\) \( = \frac{5}{3}.\)
b) Cách 1: Đặt \(x = \sin t\) ta có: \(dx = \cos tdt.\)
Khi \(x = 0\) thì \(t = 0\); khi \(x = 1\) thì \(t = \frac{\pi }{2}.\)
Do đó: \(\int_0^1 {\sqrt {1 – {x^2}} } dx\) \( = \int_0^{\frac{\pi }{2}} {\sqrt {1 – {{\sin }^2}t} } \cos tdt\) \( = \int_0^{\frac{\pi }{2}} | \cos t|.\cos tdt\) \( = \int_0^{\frac{\pi }{2}} {{{\cos }^2}tdt} \) (vì \(\cos t \ge 0\), \(\forall t \in \left[ {0;\frac{\pi }{2}} \right]\)).
\( \Rightarrow \int_0^1 {\sqrt {1 – {x^2}} } dx\) \( = \int_0^{\frac{\pi }{2}} {{{\cos }^2}tdt} \) \( = \frac{1}{2}\int_0^{\frac{\pi }{2}} {(1 + \cos 2t)dt} .\)
\( = \frac{1}{2}\int_0^{\frac{\pi }{2}} {dt} + \frac{1}{4}\int_0^{\frac{\pi }{2}} {\cos 2td(2t)} \) \( = \left. {\frac{1}{2}t} \right|_0^{\frac{\pi }{2}} + \left. {\frac{1}{4}\sin 2t} \right|_0^{\frac{\pi }{2}}\) \( = \frac{\pi }{4} + 0 = \frac{\pi }{4}.\)
Vậy \(\int_0^1 {\sqrt {1 – {x^2}} } dx = \frac{\pi }{4}.\)
Cách 2: Ta có \(y = \sqrt {1 – {x^2}} \), \(0 \le x \le 1\) là phương trình của cung của đường tròn tâm \(O\), bán kính \(R = 1.\)

Vậy tích phân \(\int_0^1 {\sqrt {1 – {x^2}} } dx\) bằng diện tích hình phẳng giới hạn bởi cung \(AB\) và hai trục tọa độ. Hình phẳng đó là một phần tư đường tròn bán kính \(1\), do đó: \(\int_0^1 {\sqrt {1 – {x^2}} } dx = \frac{\pi }{4}.\)
c) Đặt \(u = x{e^x}\) ta có: \(du = \left( {{e^x} + x{e^x}} \right)dx\) \( = {e^x}(x + 1)dx.\)
Khi \(x = 0\) thì \(u = 0\); khi \(x = 1\) thì \(u = e.\)
Do đó: \(\int_0^1 {\frac{{{e^x}(x + 1)}}{{1 + x{e^x}}}dx} \) \( = \int_0^e {\frac{{du}}{{1 + u}}} \) \( = \left. {\ln (1 + u)} \right|_0^e\) \( = \ln (e + 1).\)
d) Đặt \(x = a\sin t\) ta có: \(dx = a\cos tdt.\)
Khi \(x = 0\) thì \(t = 0\); khi \(x = \frac{a}{2}\) thì \(t = \frac{\pi }{6}.\)
Do đó: \(\int_0^{\frac{a}{2}} {\frac{1}{{\sqrt {{a^2} – {x^2}} }}dx} \) \( = \int_0^{\frac{\pi }{6}} {\frac{{a\cos tdt}}{{\sqrt {{a^2} – {a^2}{{\sin }^2}t} }}} \) \( = \int_0^{\frac{\pi }{6}} {\frac{{a\cos tdt}}{{|a\cos t|}}} .\)
Vì \(a > 0\) và \(\cos t \ge 0\), \(\forall t \in \left[ {0;\frac{\pi }{6}} \right]\) nên \(\int_0^{\frac{\pi }{6}} {\frac{{a\cos tdt}}{{|a.\cos t|}}} \) \( = \int_0^{\frac{\pi }{6}} d t = \left. t \right|_0^{\frac{\pi }{6}} = \frac{\pi }{6}.\)
Vậy \(\int_0^{\frac{a}{2}} {\frac{1}{{\sqrt {{a^2} – {x^2}} }}dx} = \frac{\pi }{6}.\)

Bài 4. Sử dụng phương pháp tích phân từng phần hãy tính:
a) \(\int_0^{\frac{\pi }{2}} {(x + 1)} \sin xdx.\)
b) \(\int_1^e {{x^2}} \ln xdx.\)
c) \(\int_0^1 {\ln } (1 + x)dx.\)
d) \(\int_0^1 {\left( {{x^2} – 2x – 1} \right){e^{ – x}}dx} .\)

Lời giải:
a) Đặt \(\left\{ {\begin{array}{*{20}{l}}
{u = x + 1}\\
{dv = \sin xdx}
\end{array}} \right.\) \( \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{du = dx}\\
{v = – \cos x}
\end{array}} \right..\)
Áp dụng công thức tính tích phân từng phần ta có:
\(\int_0^{\frac{\pi }{2}} {(x + 1)} \sin xdx\) \( = – \left. {(x + 1)\cos x} \right|_0^{\frac{\pi }{2}} + \int_0^{\frac{\pi }{2}} {\cos xdx} .\)
\( = – \left[ {\left( {\frac{\pi }{2} + 1} \right)\cos \frac{\pi }{2} – \cos 0} \right] + \left. {\sin x} \right|_0^{\frac{\pi }{2}}\) \( = 2.\)
b) Đặt \(\left\{ {\begin{array}{*{20}{l}}
{u = \ln x}\\
{dv = {x^2}dx}
\end{array}} \right.\) \( \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{du = \frac{{dx}}{x}}\\
{v = \frac{{{x^3}}}{3}}
\end{array}} \right..\)
Áp dụng công thức tính tích phân từng phần ta có:
\(\int_1^e {{x^2}} \ln xdx\) \( = \left. {\frac{{{x^3}}}{3}\ln x} \right|_1^e – \frac{1}{3}\int_1^e {{x^2}} dx\) \( = \frac{{{e^3}}}{3} – \left. {\frac{1}{9}{x^3}} \right|_1^e\) \( = \frac{{{e^3}}}{3} – \frac{1}{9}\left( {{e^3} – 1} \right)\) \( = \frac{2}{9}{e^3} + \frac{1}{9}.\)
c) Đặt \(\left\{ {\begin{array}{*{20}{l}}
{u = \ln (1 + x)}\\
{dv = dx}
\end{array}} \right.\) \( \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{du = \frac{1}{{1 + x}}dx}\\
{v = x}
\end{array}} \right.\)
Ta có: \(\int_0^1 {\ln } (1 + x)dx\) \( = \left. {x\ln (1 + x)} \right|_0^1 – \int_0^1 {\frac{x}{{1 + x}}dx.} \)
\( = \ln 2 – \int_0^1 {\frac{{x + 1 – 1}}{{1 + x}}dx} \) \( = \ln 2 – \int_0^1 d x + \int_0^1 {\frac{{dx}}{{1 + x}}} .\)
\( = \ln 2 – \left. x \right|_0^1 + \left. {\ln (1 + x)} \right|_0^1\) \( = \ln 2 – 1 + \ln 2\) \( = 2\ln 2 – 1.\)
d) Đặt \(\left\{ {\begin{array}{*{20}{l}}
{u = {x^2} – 2x – 1}\\
{dv = {e^{ – x}}dx}
\end{array}} \right.\)\( \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{du = (2x – 2)dx}\\
{v = – {e^{ – x}}}
\end{array}} \right..\)
Áp dụng công thức tính tích phân từng phần ta có:
\(\int_0^1 {\left( {{x^2} – 2x – 1} \right){e^{ – x}}dx} \) \( = – \left. {{e^{ – x}}\left( {{x^2} – 2x – 1} \right)} \right|_0^1\) \( + \int_0^1 {(2x – 2){e^{ – x}}dx} \) \( = \frac{2}{e} – 1 + 2\int_0^1 {(x – 1){e^{ – x}}dx.} \)
Tiếp tục đặt: \(\left\{ {\begin{array}{*{20}{l}}
{{u_1} = x – 1}\\
{d{v_1} = {e^{ – x}}dx}
\end{array}} \right.\) \( \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{d{u_1} = du}\\
{{v_1} = – {e^{ – x}}}
\end{array}} \right..\)
Ta có: \(\int_0^1 {(x – 1){e^{ – x}}dx} \) \( = – \left. {{e^{ – x}}(x – 1)} \right|_0^1 + \int_0^1 {{e^{ – x}}} dx\) \( = – 1 – \left. {{e^{ – x}}} \right|_0^1\) \( = – 1 – \frac{1}{e} + 1\) \( = – \frac{1}{e}.\)
Vậy \(\int_0^1 {\left( {{x^2} – 2x – 1} \right){e^{ – x}}dx} \) \( = \frac{2}{e} – 1 – \frac{2}{e} = – 1.\)

Bài 5. Tính các tích phân sau:
a) \(\int_0^1 {{{(1 + 3x)}^{\frac{3}{2}}}dx} .\)
b) \(\int_0^{\frac{1}{2}} {\frac{{{x^3} – 1}}{{{x^2} – 1}}dx.} \)
c) \(\int_1^2 {\frac{{\ln (1 + x)}}{{{x^2}}}dx} .\)

Lời giải:
a) Đặt \(u = 1 + 3x\) ta có: \(du = 3dx.\)
Khi \(x = 0\) thì \(u = 1\); khi \(x = 1\) thì \(u = 4.\)
Do đó: \(\int_0^1 {{{(1 + 3x)}^{\frac{3}{2}}}} dx\) \( = \frac{1}{3}\int_1^4 {{u^{\frac{3}{2}}}} du\) \( = \left. {\frac{1}{3}.\frac{2}{5}{u^{\frac{5}{2}}}} \right|_1^4\) \( = \frac{2}{{15}}\left( {{4^{\frac{5}{2}}} – 1} \right)\) \( = \frac{{62}}{{15}}.\)
b) \(\int_0^{\frac{1}{2}} {\frac{{{x^3} – 1}}{{{x^2} – 1}}dx} \) \( = \int_0^{\frac{1}{2}} {\frac{{(x – 1)\left( {{x^2} + x + 1} \right)}}{{(x – 1)(x + 1)}}dx} \) \( = \int_0^{\frac{1}{2}} {\frac{{{x^2} + x + 1}}{{x + 1}}dx} \) \( = \int_0^{\frac{1}{2}} {\left( {x + \frac{1}{{x + 1}}} \right)dx.} \)
\( = \left. {\frac{{{x^2}}}{2}} \right|_0^{\frac{1}{2}} + \left. {\ln (x + 1)} \right|_0^{\frac{1}{2}}\) \( = \frac{1}{8} + \ln \frac{3}{2}.\)
c) Đặt \(\left\{ {\begin{array}{*{20}{l}}
{u = \ln (1 + x)}\\
{dv = \frac{{dx}}{{{x^2}}}}
\end{array}} \right.\) \( \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{du = \frac{1}{{1 + x}}dx}\\
{v = – \frac{1}{x}}
\end{array}} \right..\)
Ta có: \(\int_1^2 {\frac{{\ln (1 + x)}}{{{x^2}}}dx} \) \( = – \left. {\frac{1}{x}\ln (1 + x)} \right|_1^2 + \int_1^2 {\frac{{dx}}{{x(1 + x)}}} \) \( = – \frac{1}{2}\ln 3 + \ln 2 + \int_1^2 {\frac{{dx}}{{x(1 + x)}}} .\)
Xét \(\int_1^2 {\frac{{dx}}{{(x + 1)x}}} .\) Ta có: \(\frac{1}{{x(1 + x)}} = \frac{1}{x} – \frac{1}{{x + 1}}.\)
Do đó: \(\int_1^2 {\frac{{dx}}{{x(x + 1)}}} \) \( = \int_1^2 {\left( {\frac{1}{x} – \frac{1}{{x + 1}}} \right)dx} \) \( = \int_1^2 {\frac{{dx}}{x}} – \int_1^2 {\frac{{dx}}{{x + 1}}} \) \( = \left. {\ln x} \right|_1^2 – \left. {\ln (1 + x)} \right|_1^2.\)
\( = \ln 2 – \ln 3 + \ln 2.\)
Vậy \(\int_1^2 {\frac{{\ln (1 + x)}}{{{x^2}}}dx} \) \( = – \frac{1}{2}\ln 3 + \ln 2 + \ln 2 – \ln 3 + \ln 2.\)
\( = 3\ln 2 – \frac{3}{2}\ln 3\) \( = \ln 8 – \frac{3}{2}\ln 3\) \( = 3\ln \frac{{2\sqrt 3 }}{3}.\)

Bài 6. Tính \(\int_0^1 {{x^2}} {(1 – x)^5}dx\) bằng hai phương pháp:
a) Đổi biến số \(u = 1 – x.\)
b) Tích phân từng phần.

Lời giải:
a) Đặt \(u = 1 – x\) ta có: \(du = – dx.\)
Đổi cận:

Ta có: \(x = 1 – u\) nên:
\(J = \int_1^0 {(u – 1){u^5}du} \) \( = \int_1^0 {{u^6}} du – \int_1^0 {{u^5}} du\) \( = \left. {\frac{{{u^7}}}{7}} \right|_1^0 – \left. {\frac{{{u^6}}}{6}} \right|_1^0\) \( = \frac{1}{{42}}.\)
b) \(J = – \frac{1}{6}\int_0^1 {xd} {(x – 1)^6}\) \( = – \frac{1}{6}\left[ {\left. {x{{(x – 1)}^6}} \right|_0^1 – \int_0^1 {{{(x – 1)}^6}} dx} \right].\)
\( = – \frac{1}{6}\left[ { – \int_0^1 {{{(x – 1)}^6}} d(x – 1)} \right]\) \( = \left. {\frac{1}{{42}}{{(x – 1)}^7}} \right|_0^1 = \frac{1}{{42}}.\)