Giải bài 3.19 trang 171 - SBT Đại số và Giải tích lớp 12

a) Đặt \(\left\{ \begin{aligned} & u=x+1 \\ & dv=\cos \left( \frac{\pi }{2}+x \right)dx \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=dx \\ & v=\sin \left( \frac{\pi }{2}+x \right) \\ \end{aligned} \right. \)

Ta có: 

\(\begin{aligned} & I=\left( x+1 \right)\sin \left( \frac{\pi }{2}+x \right)\left| \begin{aligned} & \frac{\pi }{2} \\ & 0 \\ \end{aligned} \right.-\int\limits_{0}^{\frac{\pi }{2}}{\sin \left( \frac{\pi }{2}+x \right)dx} \\ & \,\,\,=-1+\cos \left( \frac{\pi }{2}+x \right)\left| \begin{aligned} & \frac{\pi }{2} \\ & 0 \\ \end{aligned} \right.=-2 \\ \end{aligned} \)

b) Ta có: 

 \(\begin{aligned} & {{\log }_{2}}\left( x+1 \right)=\frac{\ln \left( x+1 \right)}{\ln 2} \\ & \Rightarrow \frac{{{x}^{2}}+x+1}{x+1}{{\log }_{2}}\left( x+1 \right)=\frac{1}{\ln 2}\left[ x\ln \left( x+1 \right)+\frac{\ln \left( x+1 \right)}{x+1} \right] \\ & \Rightarrow I=\int\limits_{0}^{1}{\frac{{{x}^{2}}+x+1}{x+1}{{\log }_{2}}\left( x+1 \right)dx}=\frac{1}{\ln 2}\left[ \int\limits_{0}^{1}{x\ln \left( x+1 \right)dx}+\int\limits_{0}^{1}{\frac{\ln \left( x+1 \right)}{x+1}dx} \right] \\ & \,\,\,\,\,\,\,\,\,=\dfrac{1}{\ln2}\left({{I}_{1}}+{{I}_{2}}\right) \\ \end{aligned} \)

+) \({{I}_{1}}=\int\limits_{0}^{1}{x\ln \left( x+1 \right)dx}\)

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

Ta có:

\(\begin{aligned} & {{I}_{1}}=\frac{1}{2}{{x}^{2}}\ln \left( x+1 \right)\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right.-\frac{1}{2}\int\limits_{0}^{1}{\frac{{{x}^{2}}}{x+1}dx} \\ & \,\,\,\,\,=\frac{1}{2}\ln 2-\frac{1}{2}\int\limits_{0}^{1}{\left( x-1+\frac{1}{x+1} \right)dx} \\ & \,\,\,\,=\frac{1}{2}\ln 2-\frac{1}{2}\left[ \frac{{{x}^{2}}}{2}-x+\ln \left( x+1 \right) \right]\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right. \\ & \,\,\,\,=\frac{1}{2}\ln 2-\frac{1}{2}\left[ \frac{1}{2}-1+\ln 2 \right] \\ & \,\,\,\,=\frac{1}{4} \\ \end{aligned} \)

+ Tính \(I_2\)

 \(\begin{aligned} & {{I}_{2}}=\int\limits_{0}^{1}{\frac{\ln \left( x+1 \right)}{x+1}}dx=\int\limits_{0}^{1}{\ln \left( x+1 \right)}d\left[ \ln \left( x+1 \right) \right] \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{2}{{\ln }^{2}}\left( x+1 \right)\left| \begin{aligned} & 1 \\ & 0 \\ \end{aligned} \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{2}{{\ln }^{2}}2 \\ \end{aligned} \)

Ta có: \(I=\dfrac{1}{\ln 2}\left( \dfrac{1}{4}+\dfrac{1}{2}{{\ln }^{2}}2 \right)=\dfrac{1}{4\ln 2}+\dfrac{1}{2}\ln 2\)

c) Ta có: \(\dfrac{{{x}^{2}}-1}{{{x}^{4}}+1}=\dfrac{1-\dfrac{1}{{{x}^{2}}}}{{{x}^{2}}+\dfrac{1}{{{x}^{2}}}}=\dfrac{1-\dfrac{1}{{{x}^{2}}}}{\left( x+\dfrac{1}{x} \right)-2}\)

Đặt \(t=x+\dfrac{1}{x}\Rightarrow dt=\left( 1-\dfrac{1}{{{x}^{2}}} \right)dx\), ta có:

\(\int\limits_{\frac{1}{2}}^{1}{\dfrac{{{x}^{2}}-1}{{{x}^{4}}+1}dx=}\int\limits_{\dfrac{5}{2}}^{2}{\dfrac{dt}{{{t}^{2}}-2}}=\dfrac{1}{2\sqrt{2}}\ln \left| \dfrac{t-\sqrt{2}}{t+\sqrt{2}} \right|\left| \begin{align} & 2 \\ & \frac{5}{2} \\ \end{align} \right.=\dfrac{1}{2\sqrt{2}}\ln \dfrac{6-\sqrt{2}}{6+\sqrt{2}} \)

d) 

\( \begin{aligned} & \frac{\sin 2x}{3+4\sin x-\cos 2x}=\frac{2\sin x\cos x}{3+4\sin x-\left( 1-2{{\sin }^{2}}x \right)}=\frac{\sin x\cos x}{{{\sin }^{2}}x+2\sin x+1}=\frac{\sin x\cos x}{{{\left( \sin x+1 \right)}^{2}}} \\ & \Rightarrow \int\limits_{0}^{\frac{\pi }{2}}{\frac{\sin 2x}{3+4\sin x-\cos 2x}dx}=\int\limits_{0}^{\frac{\pi }{2}}{\frac{\sin xd\left( \sin x+1 \right)}{{{\left( \sin x+1 \right)}^{2}}}dx} \\ & =\left( \ln \left| \sin x+1 \right|+\frac{1}{\sin x+1} \right)\left| \begin{aligned} & \frac{\pi }{2} \\ & 0 \\ \end{aligned} \right. \\ & =\ln 2-\frac{1}{2} \\ \end{aligned} \)