To solve the integral \(\int \frac{1}{(x-3)(x-7)} \, dx\), we will use partial fraction decomposition. The integrand can be decomposed as follows:
First, express \(\frac{1}{(x-3)(x-7)}\) in terms of partial fractions:
\(\frac{1}{(x-3)(x-7)} = \frac{A}{x-3} + \frac{B}{x-7}\),
where \(A\) and \(B\) are constants that we need to determine. Multiply through by the denominator \((x-3)(x-7)\) to eliminate the fractions:
\(1 = A(x - 7) + B(x - 3)\).
Expanding the equation gives:
\(1 = Ax - 7A + Bx - 3B\)
Combine like terms:
\(1 = (A + B)x - (7A + 3B)\)
By comparing coefficients, we get two equations:
From the first equation, \(B = -A\). Substitute this into the second equation:
\(-7A - 3(-A) = 1\)
This simplifies to:
\(-7A + 3A = 1 \rightarrow -4A = 1 \rightarrow A = -\frac{1}{4}\)
Substituting \(A = -\frac{1}{4}\) into \(B = -A\) gives:
\(B = \frac{1}{4}\)
Thus, the partial fraction decomposition of the integrand is:
\(\frac{1}{(x-3)(x-7)} = \frac{-\frac{1}{4}}{x-3} + \frac{\frac{1}{4}}{x-7}\)
The integral becomes:
\(\int \left(-\frac{1}{4(x-3)} + \frac{1}{4(x-7)}\right) \, dx\)
We can integrate each term separately:
\(\int -\frac{1}{4(x-3)} \, dx = -\frac{1}{4} \log |x-3| + C_1\)
\(\int \frac{1}{4(x-7)} \, dx = \frac{1}{4} \log |x-7| + C_2\)
Thus, the integral evaluates to:
\(-\frac{1}{4} \log |x-3| + \frac{1}{4} \log |x-7| + C\)
Combining the terms, we have:
\(\frac{1}{4} \log \left(\frac{x-7}{x-3}\right) + C\)
Therefore, the correct option is:
\(\frac{1}{4} \log \left( \frac{x-7}{x-3} \right) + c\)
The value of : \( \int \frac{x + 1}{x(1 + xe^x)} dx \).