# Họ nguyên hàm của hàm số $y=\frac{\left(2{x}^{2}+x\right)\mathrm{ln}x+1}{x}$ là

A.$\left({x}^{2}+x+1\right)\mathrm{ln}x-\frac{{x}^{2}}{2}+x+C$ .
B.$\left({x}^{2}+x-1\right)\mathrm{ln}x+\frac{{x}^{2}}{2}-x+C$ .
C.$\left({x}^{2}+x+1\right)\mathrm{ln}x-\frac{{x}^{2}}{2}-x+C$ .
D.$\left({x}^{2}+x-1\right)\mathrm{ln}x-\frac{{x}^{2}}{2}+x+C$ .
Đáp án:C
Lời giải:Lời giải
Chọn C
Ta có: $\int \frac{\left(2{x}^{2}+x\right)\mathrm{ln}x+1}{x}\text{d}x=\int \left(2x+1\right)\mathrm{ln}x\text{\hspace{0.17em}}\text{d}x+\int \frac{1}{x}\text{d}x={I}_{1}+{I}_{2}$ .
${I}_{1}=\int \left(2x+1\right)\mathrm{ln}x\text{\hspace{0.17em}}\text{d}x$ . Đặt $\left\{\begin{array}{l}u=\mathrm{ln}x\\ \text{d}v=\left(2x+1\right)\text{d}x\end{array}\right\⇒\left\{\begin{array}{l}\text{d}u=\frac{1}{x}\text{d}x\\ v={x}^{2}+x\end{array}\right\$ .
$\begin{array}{l}{I}_{1}=\left({x}^{2}+x\right)\mathrm{ln}x-\int \left({x}^{2}+x\right)\frac{1}{x}\text{d}x=\left({x}^{2}+x\right)\mathrm{ln}x-\int \left(x+1\right)\text{d}x\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left({x}^{2}+x\right)\mathrm{ln}x-\frac{{x}^{2}}{2}-x+{C}_{1}.\end{array}$
${I}_{2}=\int \frac{1}{x}\text{d}x=\mathrm{ln}x+{C}_{2}$ .
$\begin{array}{l}\int \frac{\left(2{x}^{2}+x\right)\mathrm{ln}x+1}{x}\text{d}x={I}_{1}+{I}_{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left({x}^{2}+x\right)\mathrm{ln}x-\frac{{x}^{2}}{2}-x+{C}_{1}+\mathrm{ln}x+{C}_{2}=\left({x}^{2}+x+1\right)\mathrm{ln}x-\frac{{x}^{2}}{2}-x+C.\end{array}$ .

Vậy đáp án đúng là C.