Cho hàm số \(f\left( x \right)\) liên tục trên \(\mathbb{R}\), thỏa mãn \(f\left( x \right) = x\left( {1 + \frac{1}{{\sqrt x }} – f’\left( x \right)} \right)\,,\,\forall x \in \left( {0\,; + \infty } \right)\) và\(f\left( 4 \right) = \frac{4}{3}\). Giá trị của\(\int\limits_1^4 {\left( {{x^2} – 1} \right)f’\left( x \right){\rm{d}}x} \) bằng

Cho hàm số \(f\left( x \right)\) liên tục trên \(\mathbb{R}\), thỏa mãn \(f\left( x \right) = x\left( {1 + \frac{1}{{\sqrt x }} – f’\left( x \right)} \right)\,,\,\forall x \in \left( {0\,; + \infty } \right)\) và\(f\left( 4 \right) = \frac{4}{3}\). Giá trị của\(\int\limits_1^4 {\left( {{x^2} – 1} \right)f’\left( x \right){\rm{d}}x} \) bằng
A. \(\frac{{457}}{{15}}\). B. \(\frac{{457}}{{30}}\). C. \( – \frac{{263}}{{30}}\). D. \( – \frac{{263}}{{15}}\).

Lời giải

\(\forall x \in \left( {0\,; + \infty } \right)\), ta có:
\(f\left( x \right) = x + \sqrt x – xf’\left( x \right)\,\)\( \Leftrightarrow f\left( x \right) + xf’\left( x \right)\, = x + \sqrt x \) \( \Leftrightarrow {\left( {xf\left( x \right)} \right)^\prime }\, = x + \sqrt x \).
Suy ra:
\(\int {{{\left( {xf\left( x \right)} \right)}^\prime }{\rm{d}}x} \, = \int {\left( {x + \sqrt x } \right){\rm{d}}x} \)
\( \Leftrightarrow xf\left( x \right)\, = \frac{{{x^2}}}{2} + \frac{{2\sqrt {{x^3}} }}{3} + C\)
\( \Leftrightarrow f\left( x \right)\, = \frac{x}{2} + \frac{{2\sqrt x }}{3} + \frac{C}{x}\).
Vì\(f\left( 4 \right) = \frac{4}{3}\) nên \(C = – 8\).
Suy ra \(f\left( x \right)\, = \frac{x}{2} + \frac{{2\sqrt x }}{3} – \frac{8}{x}\)\( \Rightarrow f’\,\left( x \right)\, = \frac{1}{2} + \frac{1}{{3\sqrt x }} + \frac{8}{{{x^2}}}\).
\( \Rightarrow \int\limits_1^4 {\left( {{x^2} – 1} \right)f’\left( x \right){\rm{d}}x} = \int\limits_1^4 {\left[ {\left( {{x^2} – 1} \right)\left( {\frac{1}{2} + \frac{1}{{3\sqrt x }} + \frac{8}{{{x^2}}}} \right)} \right]{\rm{d}}x} = \frac{{457}}{{15}}\).