Đáp án:
\[\int {\frac{x}{{x + \sqrt {{x^2} - 1} }}dx} = \frac{{{x^3} - \sqrt {{{\left( {{x^2} - 1} \right)}^3}} }}{3} + C\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\int {\frac{x}{{x + \sqrt {{x^2} - 1} }}dx} \\
= \int {\frac{{x\left( {x - \sqrt {{x^2} - 1} } \right)}}{{\left( {x + \sqrt {{x^2} - 1} } \right)\left( {x - \sqrt {{x^2} - 1} } \right)}}dx} \\
= \int {\frac{{x\left( {x - \sqrt {{x^2} - 1} } \right)}}{{{x^2} - \left( {{x^2} - 1} \right)}}dx} \\
= \int {x\left( {x - \sqrt {{x^2} - 1} } \right)dx} \\
= \int {{x^2}dx} - \int {x.\sqrt {{x^2} - 1} dx} \\
t = {x^2} - 1 \Rightarrow dt = 2xdx\\
\Rightarrow \int {x\sqrt {{x^2} - 1} dx} = \int {\sqrt t .\frac{{dt}}{2}} = \frac{1}{2}.\int {{t^{\frac{1}{2}}}dt} = \frac{1}{2}.\frac{{{t^{\frac{1}{2} + 1}}}}{{\frac{1}{2} + 1}} + C\\
= \frac{1}{3}.{t^{\frac{3}{2}}} + C = \frac{1}{3}.\sqrt {{t^3}} + C = \frac{1}{3}.\sqrt {{{\left( {{x^2} - 1} \right)}^3}} + C\\
\Rightarrow \int {\frac{x}{{x + \sqrt {{x^2} - 1} }}dx} = \int {{x^2}dx} - \int {x.\sqrt {{x^2} - 1} dx} \\
= \frac{{{x^3}}}{3} - \frac{1}{3}.\sqrt {{{\left( {{x^2} - 1} \right)}^3}} + C = \frac{{{x^3} - \sqrt {{{\left( {{x^2} - 1} \right)}^3}} }}{3} + C
\end{array}\)
