Bạn nào hộ mình câu này với: $\int\limits \dfrac{x^4}{x^6+1} \, dx$

Đáp án:

Giải thích các bước giải:

$\quad \displaystyle\int\dfrac{x^4}{x^6+1}dx$

$=\displaystyle\int\dfrac{x^4-x^2+1 +x^2 -1}{x^6+1}dx$

$= \displaystyle\int\dfrac{x^4-x^2 +1}{x^6+1}dx +\displaystyle\int\dfrac{x^2}{x^6+1}dx - \displaystyle\int\dfrac{1}{x^6+1}dx$

$=\displaystyle\int\dfrac{1}{x^2 +1}dx + \displaystyle\int\dfrac{x^2}{x^6+1}dx - \dfrac12\displaystyle\int\dfrac{x^4 - x^2 +1 + x^2 - (x^4-1)}{x^6+1}dx$

$= \displaystyle\int\dfrac{1}{x^2 +1}dx + \displaystyle\int\dfrac{x^2}{x^6+1}dx - \dfrac12\displaystyle\int\dfrac{x^4 - x^2 +1}{x^6+1}dx- \dfrac12\displaystyle\int\dfrac{x^2}{x^6+1}dx+ \dfrac12\displaystyle\int\dfrac{x^2 -1}{x^4 - x^2 +1}dx$

$= \dfrac12\displaystyle\int\dfrac{1}{x^2+1}dx +\dfrac12\displaystyle\int\dfrac{x^2}{x^6 +1}dx +\dfrac12\displaystyle\int\dfrac{1 -\dfrac{1}{x^2}}{x^2 -1 +\dfrac{1}{x^2}}dx$

$=\dfrac12\displaystyle\int\dfrac{1}{x^2+1}dx +\dfrac16\displaystyle\int\dfrac{d(x^3)}{(x^3)^2 +1} +\dfrac12\displaystyle\int\dfrac{d\left(x +\dfrac{1}{x}\right)}{\left(x+\dfrac1x\right)^2 - 3}$

$= \dfrac12\arctan x + \dfrac16\arctan x^3 + \dfrac12\cdot\dfrac{1}{2\sqrt3}\ln\left|\dfrac{x+\dfrac1x -\sqrt3}{x +\dfrac1x +\sqrt3}\right| + C$

$= \dfrac12\arctan x + \dfrac16\arctan x^3 + \dfrac{1}{4\sqrt3}\ln\left|\dfrac{x^2-\sqrt3x +1}{x^2+\sqrt3x +1}\right| + C$