Toán [Lớp 12] Nguyên hàm

Ta có $$ \int \frac{1+x^{1/2}}{1+x^{1/3}}dx= \int x^{1/6} dx + \int \frac{1-x^{1/6}}{1+x^{1/3}}dx = \frac{6}{7}x^{7/6}+ \int \frac{1-x^{1/6}}{1+x^{1/3}}dx.$$
Đặt $u=x^{1/6}$ thì $u^2=x^{1/3}$ và $\frac{du}{dx}= \frac{1}{6} x^{-5/6}=\frac 16 u^{-5}$. Do đó
$$\begin{align*}
I & = \int \frac{1-x^{1/6}}{1+x^{1/3}}dx, \\
& = \int \frac{1- u}{1+u^2} \cdot \frac{du}{1/6u^{-5}}, \\
& =6 \int \frac{u^5-u^6}{1+u^2} du, \\
& = 6 \int (-u^4+u^3+u^2-u-1) du + 6 \int \frac{u+2}{u^2+1}du, \\
& = -\frac 65u^5 + \frac 32 u^4 + 2u^3- 3u^2 -6u+ 3 \ln (u^2+1) + 12 \tan^{-1} u.
\end{align*}$$