Cho hàm số y=f(x) thỏa mãn f(0)=0; f'(x)=x/(x^2+1) . Họ nguyên hàm của hàm số g(x)=4xf(x) là:

Đáp án C

Ta có: $f^{\text{'}}\left(x\right)=\frac{x}{x^2+1}\Rightarrow f\left(x\right)=\int f^{\text{'}}\left(x\right)d\text{x}=\int \frac{x\text{dx}}{x^2+1}=\frac{1}{2}\int \frac{d(x^2+1)}{x^2+1}=\frac{1}{2}\ln (x^2+1)+C$

$f\left(0\right)=0\iff \frac{1}{2}\ln 1+C=0\iff C=0\Rightarrow f\left(x\right)=\frac{1}{2}\ln (x^2+1)$

$\Rightarrow g\left(x\right)=4\text{x}f\left(x\right)=2\text{x}\ln (x^2+1)\Rightarrow \int g\left(x\right)d\text{x}=\int 2\text{x}\ln (x^2+1)d\text{x}$Đặt $t=x^2+1\Rightarrow dt=2\text{xdx}$

$\Rightarrow \int g\left(x\right)d\text{x}=\int \ln t\text{d}t=t\ln t-\int t.\frac{1}{t}dt=t\ln t-\int dt=t\ln t-t+C$ 

$=(x^2+1)\ln (x^2+1)-(x^2+1)+C$

Đặt $-1+C=c\Rightarrow \int g\left(x\right)d\text{x}=(x^2+1)\ln (x^2+1)-x^2+c$ .