Giải bài 4 trang 126 – SGK môn Giải tích lớp 12
Tính:
a) \(\int{\left( 2-x \right)\sin xdx}\);
b) \(\int{\dfrac{{{\left( x+1 \right)}^{2}}}{\sqrt{x}}dx}\);
c) \(\int{\dfrac{{{e}^{3x}}+1}{{{e}^{x}}+1}dx}\);
d) \(\int{\dfrac{1}{{{\left( \sin x+\cos x \right)}^{2}}}dx}\);
e) \(\int{\dfrac{1}{\sqrt{1+x}+\sqrt{x}}dx}\);
g) \(\int{\dfrac{1}{\left( 1+x \right)\left( 2-x \right)}dx}\).
Gợi ý:
a) Sử dụng tích phân từng phần.
b) c) Thực hiện chia đa thức.
d) Biến đổi về \(\sin (x+\dfrac \pi 4)\)
e) Trục căn thức ở mẫu.
a) Đặt \(\left\{ \begin{aligned} & 2-x=u \\ & \sin xdx=dv \\ \end{aligned} \right.\Rightarrow \left\{ \begin{aligned} & du=-dx \\ & v=-\cos x \\ \end{aligned} \right. \)
\(\begin{aligned} \int{\left( 2-x \right)\sin xdx}&=-\left( 2-x \right)\cos x-\int{\cos xdx} \\ & =\left( x-2 \right)\cos x-\sin x+C \\ \end{aligned} \)
b)
\(\begin{aligned} \int{\dfrac{{{\left( x+1 \right)}^{2}}}{\sqrt{x}}dx} &=\int{\left( {{x}^{\frac{3}{2}}}+2{{x}^{\frac{1}{2}}} +{{x}^{-\frac{1}{2}}} \right)dx} \\ & =\dfrac{2}{5}{{x}^{\frac{5}{2}}}+\dfrac{4}{3}{{x}^{\frac{3}{2}}}+2{{x}^{\frac{1}{2}}}+C \\ \end{aligned}\)
c)
\(\begin{aligned} \int{\dfrac{{{e}^{3x}}+1}{{{e}^{x}}+1}dx}&=\int{\left( {{e}^{2x}}-{{e}^{x}}+1 \right)dx} \\ & =\dfrac{{{e}^{2x}}}{2}-{{e}^{x}}+x+C \\ \end{aligned} \)
d)
\(\begin{aligned} \int{\dfrac{1}{{{\left( \sin x+\cos x \right)}^{2}}}dx}&=\int{\dfrac{1}{2{{\sin }^{2}}\left( x+\dfrac{\pi }{4} \right)}dx} \\ & =-\dfrac{1}{2}\cot \left( x+\dfrac{\pi }{4} \right)+C \\ \end{aligned} \)
e)
\(\begin{aligned} \int{\dfrac{1}{\sqrt{1+x}+\sqrt{x}}dx} &=\int{\left( \sqrt{1+x}-\sqrt{x} \right)dx} \\ & =\int{\left[ {{\left( 1+x \right)}^{\frac{1}{2}}}-{{\left( x \right)}^{\frac{1}{2}}} \right]dx} \\ & =\dfrac{2}{3}{{\left( 1+x \right)}^{\frac{3}{2}}}-\dfrac{2}{3}{{x}^{\frac{3}{2}}}+C \\ \end{aligned} \)
g)
\(\begin{aligned} \int{\dfrac{1}{\left( 1+x \right)\left( 2-x \right)}dx} &=\dfrac{1}{3}\int{\left( \dfrac{1}{1+x}+\dfrac{1}{2-x} \right)dx} \\ & =\dfrac{1}{3}\ln \left| \dfrac{ 1+x }{2-x } \right|+C \\ \end{aligned} \)