Tìm họ nguyên hàm của hàm số $f(x) = {\tan ^5}x$.

$I = \int {f(x)dx}  = \int {{{\tan }^5}xdx} $.

Đặt $\tan {\mkern 1mu} x = t \Rightarrow \dfrac{{dx}}{{{{\cos }^2}x}} = dt \Rightarrow ({\tan ^2}x + 1)dx = dt \Rightarrow dx = \dfrac{{dt}}{{{t^2} + 1}}$

Khi đó:

\(\begin{array}{*{20}{l}}{I = \int {{t^5}.\dfrac{{dt}}{{{t^2} + 1}}}  = \int {({t^3} - t + \dfrac{t}{{{t^2} + 1}})dt}  = \int {{t^3}dt}  - \int {tdt}  + \int {\dfrac{t}{{{t^2} + 1}}dt} }\\{ = \dfrac{1}{4}{t^4} - \dfrac{1}{2}{t^2} + \dfrac{1}{2}\int {\dfrac{{d({t^2} + 1)}}{{{t^2} + 1}}}  = \dfrac{1}{4}{t^4} - \dfrac{1}{2}{t^2} + \dfrac{1}{2}\ln \left| {{t^2} + 1} \right| + C}\\{ = \dfrac{1}{4}{{\tan }^4}x - \dfrac{1}{2}{{\tan }^2}x + \dfrac{1}{2}\ln \left( {{{\tan }^2}x + 1} \right) + C}\\{ = \dfrac{1}{4}{{\tan }^4}x - \dfrac{1}{2}{{\tan }^2}x + \dfrac{1}{2}\ln \left( {\dfrac{1}{{{{\cos }^2}x}}} \right) + C}\\{ = \dfrac{1}{4}{{\tan }^4}x - \dfrac{1}{2}{{\tan }^2}x - \ln \left| {\cos x} \right| + C}\end{array}\)