Biết $\displaystyle\displaystyle\int\limits_{-1}^1\left(\dfrac{9}{x-3}-\dfrac{7}{x-2}\right)\mathrm{\,d}x=a\ln{3}-b\ln{2}$. Tính giá trị $P=a^2+b^2$.
 | $P=32$ |
 | $P=130$ |
 | $P=2$ |
 | $P=16$ |
Chọn phương án B.
Ta có $a\ln{3}-b\ln{2}=\ln3^a-\ln2^b=\ln\dfrac{3^a}{2^b}$.
Lại có $\displaystyle\displaystyle\int\limits_{-1}^1\left(\dfrac{9}{x-3}-\dfrac{7}{x-2}\right)\mathrm{\,d}x\approx1,451961\ldots$
Khi đó $\ln\dfrac{3^a}{2^b}\approx1,451961\ldots$ nên $$\dfrac{3^a}{2^b}=\dfrac{2187}{512}=\dfrac{3^7}{2^9}$$
Vậy $a=7$, $b=9$. Do đó $P=7^2+9^2=130$.
Chọn phương án B.
$\begin{aligned}\displaystyle\int\limits_{-1}^1\left(\dfrac{9}{x-3}-\dfrac{7}{x-2}\right)\mathrm{\,d}x&=\left(9\ln|x-3|-7\ln|x-2|\right)\bigg|_{-1}^1\\ &=7\ln3-9\ln2.\end{aligned}$
Suy ra $\begin{cases}
a=7\\ b=9
\end{cases}$. Vậy $P=a^2+b^2=130$.