Đáp án:
$I=\dfrac{1}{12}\left({\dfrac{1}{x^4}+\dfrac 13\ln\left|{\dfrac{3}{x^4}-1}\right|}\right)+C$
Giải thích các bước giải:
$I=\displaystyle\int \dfrac{1}{x^9-3x^5}dx$
$=\displaystyle\int \dfrac{1}{x^5(x^4-3)}dx$
$=\displaystyle\int \dfrac{1}{x^4-3}d\left({\dfrac{-1}{4x^4}}\right)$
$=-\dfrac 14\displaystyle\int \dfrac{1}{x^4-3}d\left({\dfrac{1}{x^4}}\right)$
Đặt $\dfrac{1}{x^4}=t$
$\to x^4=\dfrac 1t$
$\to I=-\dfrac 14\displaystyle\int \dfrac{1}{\frac{1}{t}-3}dt$
$\to I=-\dfrac 14\displaystyle\int \dfrac{t}{1-3t}dt$
$\to I=\dfrac 14\displaystyle\int \dfrac{t}{3t-1}dt$
$\to I=\dfrac{1}{12}\displaystyle\int \dfrac{3t}{3t-1}dt$
$\to I=\dfrac{1}{12}\displaystyle\int 1+\dfrac{1}{3t-1}dt$
$\to I=\dfrac{1}{12}\displaystyle\int 1+\dfrac 13\dfrac{3}{3t-1}dt$
$\to I=\dfrac{1}{12}\left({t+\dfrac 13\ln|3t-1|}\right)+C$
$\to I=\dfrac{1}{12}\left({\dfrac{1}{x^4}+\dfrac 13\ln\left|{\dfrac{3}{x^4}-1}\right|}\right)+C$
