If $\int \frac{x^{2} - 8 x + 7}{{(x^{2} - 3 x - 10)}^{2}} d x = P \log_{e} | x - 5 | + Q . \frac{1}{x - 5} - R . \log_{e} | x + 2 | - s . \frac{1}{x + 2} + c$ , then

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R = - \frac{15}{49}

S = \frac{27}{49}

P = - \frac{45}{98}

Q = \frac{8}{49}

answer is D.

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Detailed Solution

\int \frac{x^{2} - 8 x + 7}{{(x^{2} - 3 x - 10)}^{2}} d x = \int \frac{(x - 1) (x - 7)}{{(x - 5)}^{2} {(x + 2)}^{2}} d x

$\frac{(x - 1) (x - 7)}{{(x - 5)}^{2} {(x + 2)}^{2}} = \frac{A}{x - 5} + \frac{B}{{(x - 5)}^{2}} + \frac{C}{x + 2} + \frac{D}{{(x + 2)}^{2}}$

$(x - 1) (x - 7) = A (x - 5) {(x + 2)}^{2} + B {(x + 2)}^{2} + C (x + 2) {(x - 5)}^{2} + D {(x - 5)}^{2}$

Put $x = 5, B = \frac{- 8}{49}$

Put $x = - 2$

$D = \frac{27}{49}$

Coefficient of $x^{3} \Rightarrow A + C = 0$ ,

coefficient $x^{2}$ : $- A + B - 8 C + D = 1$

$- A - 8 C = 1 + \frac{8}{49} - \frac{27}{49}$

$= \frac{49 + 8 - 27}{49} = \frac{57 - 27}{49} = \frac{30}{49}$

$A + 8 C = - \frac{30}{49}$

$- C + 8 C = \frac{- 30}{49} \Rightarrow 7 C = \frac{- 30}{49}$

$C = \frac{- 30}{343}$

$∴ Q = \frac{8}{49}; S = \frac{27}{49}$

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