Nội dung bài giảng
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) \(\int_{1}^{2}\frac{ln(1+x)}{x^{2}}dx\)
Hướng dẫn giải :
a) \(\int_{0}^{1}(1+3x)^{\frac{3}{2}}dx =\frac{1}{3}\int_{0}^{1}(1+3x)^{\frac{3}{2}}d(1+3x)\)
\(=\frac{1}{3}\frac{2}{5}(1+3x)^{\frac{5}{2}}|_{0}^{1}=4\tfrac{2}{15}\)
b) \(\int_{0}^{\frac{1}{2}}\frac{x^{3}-1}{x^{2}-1}dx= \int_{0}^{\frac{1}{2}}\frac{(x-1)(x^{2}+x+1)}{(x-1)(x+1)}dx= \int_{0}^{\frac{1}{2}}\frac{x(x+1)+1}{x+1}dx\)
\(=\int_{0}^{\frac{1}{2}}(x+\frac{1}{x+1})dx=(\frac{x^{2}}{2}+ln\left | x+1 \right |)|_{0}^{\frac{1}{2}}=\frac{1}{8}+ln\frac{3}{2}\)
c) Đặt \(u = ln(1+x)\), \(dv=\frac{1}{x^{2}}dx\)\( \Rightarrow du=\frac{1}{1+x},v=-\frac{1}{x}\)
Khi đó :
\(\int_{1}^{2}\frac{ln(1+x)}{x^{2}}dx = -\frac{1}{x}ln(1+x)|_{1}^{2}+\int_{1}^{2}\frac{dx}{x(1+x)}\)
\(= - {{\ln 3} \over 2} + \ln 2 +\int\limits_1^2 {\left( {{1 \over x} - {1 \over {x + 1}}} \right)dx} \)
\(={\ln {2 \over {\sqrt 3 }} + {\rm{[}}\ln |x| - ln|x + 1|{\rm{]}}\left| {_1^2 = \ln {2 \over {\sqrt 3 }} + \ln {4 \over 3} } \right.}\)
\(= \ln {8 \over {3\sqrt 3 }}\)