Tính nguyên hàm: ∫(x² – 1) / (√x + 1) dx ∫x³ / (x – √(x²-1) ) dx

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1. 1) $\displaystyle\int\dfrac{x^2 -1}{\sqrt x +1}dx$

   $=\displaystyle\int\dfrac{(\sqrt x+1)(\sqrt x -1)(x +1)}{\sqrt x +1}dx$

   $= \displaystyle\int[(\sqrt x -1)(x+1)]dx$

   $=\displaystyle\int(x\sqrt x – x +\sqrt x -1)dx$

   $=\displaystyle\int x\sqrt xdx – \displaystyle\int xdx + \displaystyle\int\sqrt xdx – \displaystyle\int dx$

   $= \dfrac{2\sqrt{x^5}}{5} + \dfrac{x^2}{2} + \dfrac{2\sqrt{x^3}}{3} – x + C$

   2) $\displaystyle\int\dfrac{x^3}{x -\sqrt{x^2-1}}dx$

   $= \displaystyle\int\dfrac{x^3(x +\sqrt{x^2 -1})}{(x -\sqrt{x^2-1})(x +\sqrt{x^2 -1})}dx$

   $=\displaystyle\int[x^3(x +\sqrt{x^2 -1})]dx$

   $=\displaystyle\int(x^4 + x^3\sqrt{x^2 -1})dx$

   $= \displaystyle\int x^4dx + \displaystyle\int x^3\sqrt{x^2 -1}dx$

   Đặt $u = x^2$

   $\to du = 2xdx$

   Ta được:

   $\displaystyle\int x^3\sqrt{x^2 -1}dx$

   $=\dfrac12\displaystyle\int u\sqrt{u -1}du$

   Đặt $t = u -1$

   $\to dt = du$

   Ta đươc:

   $=\dfrac12\displaystyle\int (t +1)\sqrt tdt$

   $=\dfrac12\displaystyle\int (t\sqrt t + \sqrt t)dt$

   $=\dfrac12\displaystyle\int t\sqrt tdt + \dfrac12\displaystyle\int\sqrt tdt$

   Do đó, ta được:

   $\displaystyle\int\dfrac{x^3}{x -\sqrt{x^2-1}}dx$

   $= \displaystyle\int x^4dx + \dfrac12\displaystyle\int t\sqrt tdt + \dfrac12\displaystyle\int\sqrt tdt$

   $=\dfrac{x^5}{5} + \dfrac12\cdot\dfrac{2\sqrt{t^5}}{5} + \dfrac12\cdot\dfrac{2\sqrt{t^3}}{3} + C$

   $= \dfrac{x^5}{5} + \dfrac{\sqrt{t^5}}{5} + \dfrac{\sqrt{t^3}}{3} + C$

   $= \dfrac{x^5}{5} + \dfrac{\sqrt{(u-1)^5}}{5} + \dfrac{\sqrt{(u-1)^3}}{3} + C$

   $= \dfrac{x^5}{5} + \dfrac{\sqrt{(x^2-1)^5}}{5} + \cdot\dfrac{\sqrt{(x^2-1)^3}}{3} + C$

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