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

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:

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

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

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

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

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

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