|
|
Nhắc lại không chứng minh kết quả sau
$\int\limits\dfrac{1}{1-z^2}dz=\dfrac{1}{2}\ln\left| {\dfrac{1+z}{1-z}} \right|+C$
Ta có
$\dfrac{x^{2}+1}{x\sqrt{3x+1}}=\dfrac{x^{2}}{x\sqrt{3x+1}}+\dfrac{1}{x\sqrt{3x+1}}$
$\dfrac{x^{2}+1}{x\sqrt{3x+1}}=\dfrac{x}{\sqrt{3x+1}}+\dfrac{1}{x\sqrt{3x+1}}$
$\dfrac{x^{2}+1}{x\sqrt{3x+1}}=\dfrac{(3x-2)+2(3x+1)}{9\sqrt{3x+1}}-\dfrac{3}{1-(3x+1)}.\dfrac{1}{\sqrt{3x+1}}$
$\dfrac{x^{2}+1}{x\sqrt{3x+1}}=\dfrac{(3x-2)}{9\sqrt{3x+1}}+\dfrac{2(3x+1)}{9\sqrt{3x+1}}-\dfrac{1}{1-(3x+1)}.\dfrac{3}{\sqrt{3x+1}}$
$\dfrac{x^{2}+1}{x\sqrt{3x+1}}=\dfrac{2}{27}(3x-2)\left ( \sqrt{3x+1} \right )'+\dfrac{2}{27}(3x-2)'\left ( \sqrt{3x+1} \right )-2.\dfrac{1}{1-(3x+1)}\left ( \sqrt{3x+1} \right )'$
$\dfrac{x^{2}+1}{x\sqrt{3x+1}}=\left[ {\dfrac{2}{27}(3x-2)\left ( \sqrt{3x+1}
\right )} \right]'-2.\dfrac{1}{1-(3x+1)}\left ( \sqrt{3x+1} \right )'$
Suy ra
$\int\limits_{1}^{5}\dfrac{x^{2}+1}{x\sqrt{3x+1}}dx=\left[ {\dfrac{2}{27}(3x-2)\left ( \sqrt{3x+1}
\right )} \right]_{1}^{5}-2\int\limits_{1}^{5}\dfrac{1}{1-(3x+1)}d\left ( \sqrt{3x+1} \right )$
$\int\limits_{1}^{5}\dfrac{x^{2}+1}{x\sqrt{3x+1}}dx=\left[ {\dfrac{2}{27}(3x-2)\left ( \sqrt{3x+1}
\right )} \right]_{1}^{5}-2\left[ {\dfrac{1}{2}\ln\left| {\dfrac{1+\sqrt{3x+1}
}{1-\sqrt{3x+1}
}} \right|} \right]_{1}^{5}$
$\boxed{\int\limits_{1}^{5}\dfrac{x^{2}+1}{x\sqrt{3x+1}}dx=\dfrac{100}{27}+\ln\dfrac{9
}{5}}$
|