Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
1)\quad D = \displaystyle\int\limits_0^1\dfrac{2x}{x^2 - 8x + 7}dx\\
\to D = \displaystyle\int\limits_0^1\left(\dfrac{7}{3(x-7)} - \dfrac{1}{3(x-1)}\right)dx\\
\to D = \dfrac73\displaystyle\int\limits_0^1\dfrac{1}{x-7}dx - \dfrac13\displaystyle\int\limits_0^1\dfrac{1}{x-1}dx\\
\to D = \dfrac73\ln|x-7|\Bigg|_0^1 - \dfrac13\ln|x-1|\Bigg|_0^1\\
\to D = \dfrac73\left(\ln 6 - \ln 7\right) - \dfrac13\left(\ln 0 - \ln 1\right)\\
\to D = +\infty\quad \text{(phân kỳ)}\\
2)\quad E = \displaystyle\int\limits_0^2\dfrac{x+1}{x^2 - 7x + 12}dx\\
\to E = \displaystyle\int\limits_0^2\left(\dfrac{5}{x-4} - \dfrac{4}{x-3}\right)dx\\
\to E = 5\displaystyle\int\limits_0^2\dfrac{1}{x-4}dx - 4\displaystyle\int\limits_0^2\dfrac{1}{x-3}dx\\
\to E = 5\ln|x-4|\Bigg|_0^2 - 4\ln|x-3|\Bigg|_0^2\\
\to E = 5\left(\ln2 - \ln4\right) - 4\left(\ln1 - \ln3\right)\\
\to E = -5\ln2 + 4\ln3
\end{array}\)
