Biết nguyên hàm của (ln(x+3))dx = xln(x+3)+ ax+ b.ln(x+3) + C

* Hướng dẫn giải

Đáp án B

Đặt $\left\{\begin{array}{l}u=\ln \left(x+3\right)\\ \text{d}v=\text{d}x\end{array}\right.\Rightarrow \left\{\begin{array}{l}\text{d}u=\frac{1}{x+3}\text{d}x\\ v=x\end{array}\right.$

Khi đó:

$\begin{array}{l}\int \ln \left(x+3\right)\text{\hspace{0.17em}d}x\\ =x\ln \left(x+3\right)-\int \frac{x}{x+3}\text{d}x\end{array}$

$=x\ln \left(x+3\right)-\int \left(1-\frac{3}{x+3}\right)\text{d}x=x\ln \left(x+3\right)-x+3\ln \left(x+3\right)+C$

Suy ra: $\left\{\begin{array}{l}a=-1\\ b=3\end{array}\right.$. Vậy $S=2a-b=-5$ .