Cho hàm số f(x) có và \(f'\left( x \right) = \sqrt {{{\ln }^2}x + 1} .\frac{{\ln x}}{x}\) với x > 0. Khi đó bằng

* Hướng dẫn giải

Xét \(\int {f'\left( x \right)} .{\rm{d}}x = \int {\sqrt {l{n^2}x + 1} .\frac{{lnx}}{x}} .{\rm{d}}x\).

Đặt \(\sqrt {l{n^2}x + 1} = t \Rightarrow l{n^2}x = {t^2} - 1 \Rightarrow \frac{{lnx}}{x}.{\rm{d}}x = t.{\rm{d}}t\).

Suy ra: \(\int {f'\left( x \right){\rm{d}}x} = \int {t.t{\rm{d}}t} = \frac{{{t^3}}}{3} + C = \frac{{\sqrt {{{\left( {{{\ln }^2}x + 1} \right)}^3}} }}{3} + C\)

Vì vậy: \(f\left( x \right) = \frac{{\sqrt {{{\left( {l{n^2}x + 1} \right)}^3}} }}{3} + C\).

Do \(f\left( 1 \right) = \frac{1}{3} \Leftrightarrow \frac{1}{3} + C = \frac{1}{3} \Leftrightarrow C = 0\). Suy ra: \(f\left( x \right) = \frac{{\sqrt {{{\left( {l{n^2}x + 1} \right)}^3}} }}{3}\).

Vậy \(\int\limits_1^2 {\frac{{f\left( x \right)}}{{x\sqrt {{{\ln }^2}x + 1} }}dx} = \int\limits_1^2 {\frac{{\sqrt {{{({{\ln }^2}x + 1)}^3}} }}{{3x\sqrt {{{\ln }^2}x + 1} }}dx} = \int\limits_1^2 {\frac{{{{\ln }^2}x + 1}}{{3x}}dx} = \frac{1}{3}\int\limits_1^2 {\left( {{{\ln }^2}x + 1} \right)d(\ln x)}\)

\( = \left. {\frac{1}{3}\left( {\frac{1}{3}{{\ln }^3}x + \ln x} \right)} \right|_1^2 = \frac{1}{3}\left( {\frac{1}{3}{{\ln }^3}2 + \ln 2} \right) = \frac{{\ln 2\left( {{{\ln }^2}2 + 3} \right)}}{9}\).