Nguyên hàm của x^5/x^2+1 Help me

Đáp án:

$\dfrac12\cdot\left(\dfrac{x^4}{2} -x^2 + \ln|x^2 +1|\right) + C$

Giải thích các bước giải:

$\displaystyle\int\dfrac{x^5}{x^2 +1}dx$

Đặt $u = x^2$

$\to du = 2xdx$

Ta được:

$\dfrac12\displaystyle\int\dfrac{u^2}{u+1}du$

$= \dfrac12\displaystyle\int\left(u - 1 +\dfrac{1}{u+1}\right)du$

$=\dfrac12\displaystyle\int udu - \dfrac12\displaystyle\int du +\dfrac12\displaystyle\int\dfrac{du}{u+1}$

Đặt $t = u +1$

$\to dt = du$

Ta được:

$\dfrac12\displaystyle\int udu - \dfrac12\displaystyle\int du +\dfrac12\displaystyle\int\dfrac{dt}{t}$

$=\dfrac12\cdot\dfrac{u^2}{2} -\dfrac12\cdot u +\dfrac12\cdot\ln|t| + C$

$= \dfrac12\cdot\left(\dfrac{x^4}{2} - x^2 + \ln|x^2 +1|\right) + C$

Giải thích các bước giải:

Ta có:
$I=\displaystyle\int\dfrac{x^5}{x^2+1}dx$

Đặt $x^2=t\to dx^2=dt\to 2xdx=dt$

$\to I=\dfrac12\displaystyle\int\dfrac{x^4}{x^2+1}\cdot 2xdx$

$\to I=\dfrac12\displaystyle\int\dfrac{t^2}{t+1} dt$

$\to I=\dfrac12\displaystyle\int\dfrac{t^2-1+1}{t+1} dt$

$\to I=\dfrac12\displaystyle\int\dfrac{(t-1)(t+1)+1}{t+1} dt$

$\to I=\dfrac12\displaystyle\int(t-1+\dfrac{1}{t+1}) dt$

$\to I=\dfrac12(\dfrac12t^2-t+\ln|t+1|)$

$\to I=\dfrac12(\dfrac12x^4-x^2+\ln|x^2+1|)$