Giải bài 2 trang 14

Đề bài

Tính các tích phân sau:

a) $\underset{1}{\overset{2}{\int }}\frac{1-2x}{x^2}dx$;

b) $\underset{1}{\overset{2}{\int }}{\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^{2}dx$;

c) $\underset{1}{\overset{4}{\int }}\frac{x-4}{\sqrt{x}+2}dx$.

Lời giải chi tiết

a) $\underset{1}{\overset{2}{\int }}\frac{1-2x}{x^2}dx=\underset{1}{\overset{2}{\int }}\left(x^{-2}-2.\frac{1}{x}\right)dx={\left(-\frac{1}{x}-2\ln \left|x\right|\right)|}_1^2=\left(-\frac{1}{2}-2\ln 2\right)-\left(-\frac{1}{1}-2\ln 1\right)=\frac{1}{2}-2\ln 2$.

b)

$\begin{array}{c}\underset{1}{\overset{2}{\int }}{\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)}^{2}dx=\underset{1}{\overset{2}{\int }}\left(x+2+\frac{1}{x}\right)dx={\left(\frac{x^2}{2}+2x+\ln \left|x\right|\right)|}_1^2\\ =\left(\frac{2^2}{2}+2.2+\ln 2\right)-\left(\frac{1^2}{2}+2.1+\ln 1\right)=\frac{7}{2}+\ln 2\end{array}$

c)

$\begin{array}{c}\underset{1}{\overset{4}{\int }}\frac{x-4}{\sqrt{x}+2}dx=\underset{1}{\overset{4}{\int }}\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}+2}dx=\underset{1}{\overset{4}{\int }}\left(\sqrt{x}-2\right)dx=\underset{1}{\overset{4}{\int }}\left(x^{\frac{1}{2}}-2\right)dx\\ ={\left(\frac{2}{3}{x}^{\frac{3}{2}}-2x\right)|}_1^4=\left(\frac{2}{3}{.4}^{\frac{3}{2}}-2.4\right)-\left(\frac{2}{3}{.1}^{\frac{3}{2}}-2.1\right)=-\frac{4}{3}\end{array}$