Find the values of $P$ and $Q$ that make the following equality true.
$$
\begin{array}{l}
\int \frac{1-2 x}{(x+1)(x-5)} d x=P \cdot \ln |x+1|+Q \cdot \ln |x-5|+C \
P=\square \
Q=\square
\end{array}
$$

Question

Find the values of $P$ and $Q$ that make the following equality true.
$$
\begin{array}{l}
\int \frac{1-2 x}{(x+1)(x-5)} d x=P \cdot \ln |x+1|+Q \cdot \ln |x-5|+C \
P=\square \
Q=\square
\end{array}
$$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the values of $P$ and $Q$, we need to compare the given integral with the standard form of the partial fraction decomposition. ‖ ‖ step 2 ⋮ The integral $\int \frac{1-2x}{(x+1)(x-5)} dx$ can be decomposed into partial fractions as $\frac{A}{x+1} + \frac{B}{x-5}$. ‖ ‖ step 3 ⋮ To find $A$ and $B$, we solve the equation $1-2x = A(x-5) + B(x+1)$. ‖ ‖ step 4 ⋮ Setting $x=-1$ gives $A = \frac{1-2(-1)}{-1-5} = \frac{3}{-6} = -\frac{1}{2}$. ‖ ‖ step 5 ⋮ Setting $x=5$ gives $B = \frac{1-2(5)}{5+1} = \frac{-9}{6} = -\frac{3}{2}$. ‖ ‖ step 6 ⋮ The integral then becomes $\int \frac{-\frac{1}{2}}{x+1} dx + \int \frac{-\frac{3}{2}}{x-5} dx$. ‖ ‖ step 7 ⋮ Integrating gives $-\frac{1}{2} \ln|x+1| - \frac{3}{2} \ln|x-5| + C$. ‖ ‖ step 8 ⋮ Comparing with the given integral form, we find $P = -\frac{1}{2}$ and $Q = -\frac{3}{2}$. ‖ ∻Answer∻ ⚹ $P = -\frac{1}{2}$, $Q = -\frac{3}{2}$ ⚹ ∻Key Concept∻ ⚹Partial Fraction Decomposition⚹ ∻Explanation∻ ⚹To integrate a rational function, we can decompose it into simpler fractions whose integrals are known, and then integrate each term separately.⚹◊What is the value of $P$ in the equation $P \cdot \ln |x+1| + Q \cdot \ln |x-5|$?⍭ Generate me a similar question◊

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