Cho hàm số f(x) xác định và có đạo hàm f'(x) liên tục trên đoạn

* Hướng dẫn giải

$\begin{array}{l}\Rightarrow \underset{1}{\overset{x}{\int }}\frac{f\text{'}\left(x\right){\left(1+f\left(x\right)\right)}^2}{{\left(f\left(x\right)\right)}^4}dx=\underset{1}{\overset{x}{\int }}{\left(x-1\right)}^{2}dx,\forall x\in \left[1;3\right]\\ \iff \underset{1}{\overset{x}{\int }}\left[\frac{1}{{\left(f\left(x\right)\right)}^4}+\frac{2}{{\left(f\left(x\right)\right)}^3}+\frac{1}{{\left(f\left(x\right)\right)}^2}\right]d\left(f\left(x\right)\right)=\frac{{\left(x-1\right)}^3}{3}\left|\begin{array}{l}{}^x\\ {}_1\end{array}\right.\\ \iff \left[-\frac{1}{3{\left(f\left(x\right)\right)}^3}-\frac{2}{2{\left(f\left(x\right)\right)}^2}-\frac{1}{f\left(x\right)}\right]\left|\begin{array}{l}{}^x\\ {}_1\end{array}\right.=\frac{{\left(x-1\right)}^3}{3}-\frac{0}{3}\\ \iff \left[-\frac{1}{3{\left(f\left(x\right)\right)}^3}-\frac{2}{2{\left(f\left(x\right)\right)}^2}-\frac{1}{f\left(x\right)}\right]-\left[-\frac{1}{3{\left(f\left(1\right)\right)}^3}-\frac{2}{2{\left(f\left(1\right)\right)}^2}-\frac{1}{f\left(1\right)}\right]=\frac{{\left(x-1\right)}^3}{3}\\ \iff \left[-\frac{1}{3{\left(f\left(x\right)\right)}^3}-\frac{2}{2{\left(f\left(x\right)\right)}^2}-\frac{1}{f\left(x\right)}\right]-\left[\frac{1}{3}-1+1\right]=\frac{{\left(x-1\right)}^3}{3}\\ \iff -\frac{1}{3{\left(f\left(x\right)\right)}^3}-\frac{2}{2{\left(f\left(x\right)\right)}^2}-\frac{1}{f\left(x\right)}=\frac{{\left(x-1\right)}^3+1}{3}\\ \iff \frac{1}{3}{\left(-\frac{1}{f\left(x\right)}\right)}^3-{\left(-\frac{1}{f\left(x\right)}\right)}^2+\left(-\frac{1}{f\left(x\right)}\right)=\frac{1}{3}{x}^3-x^2+x(*)\end{array}$