(Sở Phú Thọ 2022) Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}3{x^2}\ln \left( {x + 1} \right) & {\rm{khi}}\,\,x \ge 0\\2x\sqrt {{x^2} + 3} + 1 & {\rm{khi}}\,\,x < 0\end{array} \right.\). Biết \(\int\limits_{\frac{1}{e}}^e {\frac{{f\left( {\ln x} \right)}}{x}{\rm{d}}x} = a\sqrt 3 + b\ln 2 + c\) với \(a,b,c \in \mathbb{Q}\). Giá trị của \(a + b + 6c\) bằng

A. \(35\).

B. \( – 14\).

C. \( – 27\).

D. \(18\).

Lời giải:

Chọn C

Ta có: \(I = \int\limits_{\frac{1}{e}}^e {\frac{{f\left( {\ln x} \right)}}{x}{\rm{d}}x} \). Đặt: \(t = \ln x \Rightarrow {\rm{d}}t = \frac{1}{x}{\rm{d}}x\). Đổi cận: \(\left\{ \begin{array}{l}x = e \Rightarrow t = 1\\x = \frac{1}{e} \Rightarrow t = – 1\end{array} \right.\).

\( \Rightarrow I = \int\limits_{ – 1}^1 {f\left( t \right){\rm{d}}t} = \int\limits_{ – 1}^1 {f\left( x \right){\rm{d}}x} = \int\limits_{ – 1}^0 {f\left( x \right){\rm{d}}x} + \int\limits_0^1 {f\left( x \right){\rm{d}}x} \)

\( \Rightarrow I = \int\limits_{ – 1}^0 {\left( {2x\sqrt {{x^2} + 3} + 1} \right){\rm{d}}x} + \int\limits_0^1 {3{x^2}\ln \left( {x + 1} \right){\rm{d}}x} \)

\( \Rightarrow I = \int\limits_{ – 1}^0 {2x\sqrt {{x^2} + 3} {\rm{d}}x} + \int\limits_{ – 1}^0 {{\rm{d}}x} + \int\limits_0^1 {3{x^2}\ln \left( {x + 1} \right){\rm{d}}x} \).

Ta có: \(J = \int\limits_{ – 1}^0 {2x\sqrt {{x^2} + 3} {\rm{d}}x} \).

Đặt: \(r = \sqrt {{x^2} + 3} \Rightarrow {r^2} = {x^2} + 3 \Rightarrow 2r{\rm{d}}r = 2x{\rm{d}}x\). Đổi cận: \(\left\{ \begin{array}{l}x = 0 \Rightarrow r = \sqrt 3 \\x = – 1 \Rightarrow r = 2\end{array} \right.\).

\( \Rightarrow J = \int\limits_2^{\sqrt 3 } {2{r^2}{\rm{d}}r} = \left. {\frac{{2{r^3}}}{3}} \right|_2^{\sqrt 3 } = 2\sqrt 3 – \frac{{16}}{3}\).

Ta có: \(K = \int\limits_0^1 {3{x^2}\ln \left( {x + 1} \right){\rm{d}}x} \).

Đặt: \(\left\{ \begin{array}{l}u = \ln \left( {x + 1} \right)\\{\rm{d}}v = 3{x^2}{\rm{d}}x\end{array} \right. \Rightarrow \left\{ \begin{array}{l}{\rm{d}}u = \frac{1}{{x + 1}}{\rm{d}}x\\v = {x^3} + 1 = \left( {x + 1} \right)\left( {{x^2} – x + 1} \right)\end{array} \right.\).

\( \Rightarrow K = \left. {\left( {{x^3} + 1} \right)\ln \left( {x + 1} \right)} \right|_0^1 – \int\limits_0^1 {\left( {{x^2} – x + 1} \right){\rm{d}}x} = 2\ln 2 – \ln 1 – \left. {\left( {\frac{{{x^3}}}{3} – \frac{{{x^2}}}{2} + 1} \right)} \right|_0^1 = 2\ln 2 – \frac{5}{6}\).

Khi đó: \(I = J + K + \int\limits_{ – 1}^0 {{\rm{d}}x} = 2\sqrt 3 – \frac{{16}}{3} + 2\ln 2 – \frac{5}{6} + 1 = 2\sqrt 3 + 2\ln 2 – \frac{{31}}{6}\).

\( \Rightarrow a = 2;\,\,b = 2;\,\,c = – \frac{{31}}{6}\). Vậy \(a + b + 6c = – 27\).