Ngân hàng bài tập
C
Tính tích phân \(I=\displaystyle\int\limits_0^1\dfrac{\mathrm{\,d}x}{x^2-9}\).
 | \(I=\dfrac{1}{6}\ln\dfrac{1}{2}\) |
 | \(I=-\dfrac{1}{6}\ln\dfrac{1}{2}\) |
 | \(I=\dfrac{1}{6}\ln2\) |
 | \(I=\ln\sqrt[6]{2}\) |
2 lời giải
Chọn phương án A.
Dùng máy tính cầm tay:
- Tính tích phân:
- Kiểm tra các phương án:
Chọn phương án A.
\(\begin{align*}
I&=\displaystyle\int\limits_0^1 \dfrac{\mathrm{\,d}x}{x^2-9}=\displaystyle\int\limits_0^1 \dfrac{\mathrm{\,d}x}{(x-3)(x+3)}\\
&=\dfrac{1}{6}\displaystyle\int\limits_0^1 \left(\dfrac{1}{x-3}-\dfrac{1}{x+3}\right)\mathrm{\,d}x\\
&=\dfrac{1}{6}\left(\ln|x-3|-\ln|x+3|\right)\bigg|_0^1\\
&=\dfrac{1}{6}\left(\ln\dfrac{2}{3}-\ln \dfrac{4}{3}\right)=\dfrac{1}{6}\ln\dfrac{1}{2}.
\end{align*}\)
Giả sử \(\dfrac{1}{(x-3)(x+3)}=\dfrac{A}{x-3}+\dfrac{B}{x+3}\)
\(\Leftrightarrow\dfrac{1}{(x-3)(x+3)}=\dfrac{Ax+3A+Bx-3B}{(x-3)(x+3)}\)
\(\Leftrightarrow\dfrac{1}{(x-3)(x+3)}=\dfrac{(A+B)x+3A-3B}{(x-3)(x+3)}\)
Khi đó \(\begin{cases}A+B&=0\\ 3A-3B&=1\end{cases}\Leftrightarrow\begin{cases}A=\dfrac{1}{6}\\ B=-\dfrac{1}{6}.\end{cases}\)
\(\begin{align*}\Rightarrow\dfrac{1}{(x-3)(x+3)}&=\dfrac{1}{6(x-3)}-\dfrac{1}{6(x+3)}\\
&=\dfrac{1}{6}\left[\dfrac{1}{x-3}-\dfrac{1}{x+3}\right]\end{align*}\)
\(\Leftrightarrow\dfrac{1}{(x-3)(x+3)}=\dfrac{Ax+3A+Bx-3B}{(x-3)(x+3)}\)
\(\Leftrightarrow\dfrac{1}{(x-3)(x+3)}=\dfrac{(A+B)x+3A-3B}{(x-3)(x+3)}\)
Khi đó \(\begin{cases}A+B&=0\\ 3A-3B&=1\end{cases}\Leftrightarrow\begin{cases}A=\dfrac{1}{6}\\ B=-\dfrac{1}{6}.\end{cases}\)
\(\begin{align*}\Rightarrow\dfrac{1}{(x-3)(x+3)}&=\dfrac{1}{6(x-3)}-\dfrac{1}{6(x+3)}\\
&=\dfrac{1}{6}\left[\dfrac{1}{x-3}-\dfrac{1}{x+3}\right]\end{align*}\)