Question

[Pham Trong Bach]

Cho hàm số y = f(x) thỏa mãn $\mathrm{f}\left(0\right)=0; \mathrm{f}\text{'}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{{\mathrm{x}}^2+1}$. Họ nguyên hàm của hàm số $\mathrm{g}\left(\mathrm{x}\right)=4\mathrm{xf}\left(\mathrm{x}\right)$là:

$\mathbf{A}\mathbf{.} ({\mathrm{x}}^2+1)\ln \left({\mathrm{x}}^2\right)-{\mathrm{x}}^2+\mathrm{c}$

$\mathbf{B}\mathbf{.} {\mathrm{x}}^2\ln ({\mathrm{x}}^2+1)-{\mathrm{x}}^2$

$\mathbf{C}\mathbf{.} ({\mathrm{x}}^2+1)\ln ({\mathrm{x}}^2+1)-{\mathrm{x}}^2+\mathrm{c}$

$\mathbf{D}\mathbf{.} ({\mathrm{x}}^2+1)\ln ({\mathrm{x}}^2+1)-{\mathrm{x}}^2$

Answer 1

[Cao Minh Tâm]