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Câu hỏi: Cho hàm số $f\left(x \right)$ liên tục trên khoảng $\left(0;+\infty \right)$ và thỏa mãn
$f\left({{x}^{2}}+1 \right)+\frac{f\left(\sqrt{x} \right)}{4x\sqrt{x}}=\frac{2x+1}{2x}.\ln \left(x+1 \right)$. Biết $\int\limits_{1}^{17}{f\left(x \right)\text{d}x=a\ln 5-2\ln b+c}$ với $a, b, c\in \mathbb{R}$. Giá trị của $a+b+2c$ bằng
A. $\frac{29}{2}$.
B. $5$.
C. $7$.
D. $37$.
$f\left({{x}^{2}}+1 \right)+\frac{f\left(\sqrt{x} \right)}{4x\sqrt{x}}=\frac{2x+1}{2x}.\ln \left(x+1 \right)$. Biết $\int\limits_{1}^{17}{f\left(x \right)\text{d}x=a\ln 5-2\ln b+c}$ với $a, b, c\in \mathbb{R}$. Giá trị của $a+b+2c$ bằng
A. $\frac{29}{2}$.
B. $5$.
C. $7$.
D. $37$.
Do $f\left( x \right)$ liên tục trên khoảng $\left( 0;+\infty \right)$ nên tồn tại $F\left( x \right)=\int{f\left( x \right)}\text{d}x$, $\forall x>0$.
Với $x>0$, ta có:
$f\left( {{x}^{2}}+1 \right)+\frac{f\left( \sqrt{x} \right)}{4x\sqrt{x}}=\frac{2x+1}{2x}.\ln \left( x+1 \right)$ $\Leftrightarrow $ $2x.f\left( {{x}^{2}}+1 \right)+\frac{f\left( \sqrt{x} \right)}{2\sqrt{x}}=\left( 2x+1 \right).\ln \left( x+1 \right)$.
Xét vế trái: $g\left( x \right)=2x.f\left( {{x}^{2}}+1 \right)+\frac{f\left( \sqrt{x} \right)}{2\sqrt{x}}$ $\Rightarrow $ $\int{g\left( x \right)}\text{d}x=F\left( {{x}^{2}}+1 \right)+F\left( \sqrt{x} \right)+{{C}_{1}}$.
Xét vế phải: $h\left( x \right)=\left( 2x+1 \right).\ln \left( x+1 \right)$ $\Rightarrow $ $\int{h\left( x \right)}\text{d}x=\int{\left( 2x+1 \right)\ln \left( x+1 \right)}\text{d}x$ $=\int{\ln \left( x+1 \right)}\text{d}\left( {{x}^{2}}+x \right)$ $=\left( {{x}^{2}}+x \right)\ln \left( x+1 \right)-\int{\left( {{x}^{2}}+x \right)\frac{1}{x+1}\text{d}x}$ $=\left( {{x}^{2}}+x \right)\ln \left( x+1 \right)-\int{x\text{d}x}$ $=\left( {{x}^{2}}+x \right)\ln \left( x+1 \right)-\frac{{{x}^{2}}}{2}+{{C}_{2}}$.
Suy ra $F\left( {{x}^{2}}+1 \right)+F\left( \sqrt{x} \right)=\left( {{x}^{2}}+x \right)\ln \left( x+1 \right)-\frac{{{x}^{2}}}{2}+C\quad \left( 1 \right)$.
Thay $x=4$ vào $\left( 1 \right)$ ta có: $F\left( 17 \right)+F\left( 2 \right)=20\ln 5-8+C$.
Thay $x=1$ vào $\left( 1 \right)$ ta có: $F\left( 2 \right)+F\left( 1 \right)=2\ln 2-\frac{1}{2}+C$.
Nên $\int\limits_{1}^{17}{f\left( x \right)\text{d}x=}F\left( 17 \right)-F\left( 1 \right)=20\ln 5-2\ln 2-\frac{15}{2}$, suy ra $a=20$, $b=2$, $c=-\frac{15}{2}$.
Vậy: $a+b+2c=20+2-15=7$. Ta chọn C.
Với $x>0$, ta có:
$f\left( {{x}^{2}}+1 \right)+\frac{f\left( \sqrt{x} \right)}{4x\sqrt{x}}=\frac{2x+1}{2x}.\ln \left( x+1 \right)$ $\Leftrightarrow $ $2x.f\left( {{x}^{2}}+1 \right)+\frac{f\left( \sqrt{x} \right)}{2\sqrt{x}}=\left( 2x+1 \right).\ln \left( x+1 \right)$.
Xét vế trái: $g\left( x \right)=2x.f\left( {{x}^{2}}+1 \right)+\frac{f\left( \sqrt{x} \right)}{2\sqrt{x}}$ $\Rightarrow $ $\int{g\left( x \right)}\text{d}x=F\left( {{x}^{2}}+1 \right)+F\left( \sqrt{x} \right)+{{C}_{1}}$.
Xét vế phải: $h\left( x \right)=\left( 2x+1 \right).\ln \left( x+1 \right)$ $\Rightarrow $ $\int{h\left( x \right)}\text{d}x=\int{\left( 2x+1 \right)\ln \left( x+1 \right)}\text{d}x$ $=\int{\ln \left( x+1 \right)}\text{d}\left( {{x}^{2}}+x \right)$ $=\left( {{x}^{2}}+x \right)\ln \left( x+1 \right)-\int{\left( {{x}^{2}}+x \right)\frac{1}{x+1}\text{d}x}$ $=\left( {{x}^{2}}+x \right)\ln \left( x+1 \right)-\int{x\text{d}x}$ $=\left( {{x}^{2}}+x \right)\ln \left( x+1 \right)-\frac{{{x}^{2}}}{2}+{{C}_{2}}$.
Suy ra $F\left( {{x}^{2}}+1 \right)+F\left( \sqrt{x} \right)=\left( {{x}^{2}}+x \right)\ln \left( x+1 \right)-\frac{{{x}^{2}}}{2}+C\quad \left( 1 \right)$.
Thay $x=4$ vào $\left( 1 \right)$ ta có: $F\left( 17 \right)+F\left( 2 \right)=20\ln 5-8+C$.
Thay $x=1$ vào $\left( 1 \right)$ ta có: $F\left( 2 \right)+F\left( 1 \right)=2\ln 2-\frac{1}{2}+C$.
Nên $\int\limits_{1}^{17}{f\left( x \right)\text{d}x=}F\left( 17 \right)-F\left( 1 \right)=20\ln 5-2\ln 2-\frac{15}{2}$, suy ra $a=20$, $b=2$, $c=-\frac{15}{2}$.
Vậy: $a+b+2c=20+2-15=7$. Ta chọn C.
Đáp án C.