Question

$\int \frac{x^2-8x+7}{{(x^2-3x-10)}^2}dx=P\log |x-5|+Q\frac{1}{x-5}-R.\log |x+2|-S.\frac{1}{x+2}+c$. Then

• A. $P=-\frac{45}{98}$
• B. $Q=\frac{8}{49}$
• C. $R=\frac{15}{49}$
• D. All of these

Answer 1

Answer: (D)

We have $\frac{x^2-8x+7}{{(x^2-3x-10)}^2}=\frac{(x-7)(x-1)}{{(x-5)}^2{(x+2)}^2}$

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

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

On putting $x=5$ and $x=-2$ respectively, we get

$-8=B{\left(7\right)}^2\Rightarrow B=-\frac{8}{49}$ and $27=D{(-7)}^2\Rightarrow D=\frac{27}{49}$

Now, on putting $4x=0$ in (1), we get

$7=-20A+4B+50C+25D\Rightarrow 7=-20A-\frac{32}{49}+50C+27\times \frac{25}{49}$

$\Rightarrow 20A-50C=\frac{300}{49}\Rightarrow 2A-5C=\frac{30}{49}$ ...(2)

On putting $x=1$ in (1), we get

$0=-16A+9B+48C+16D\Rightarrow 0=-16A+48C-\frac{72}{49}+\frac{442}{49}$

$\Rightarrow 16A-48C=\frac{360}{49}\Rightarrow 2A-6C=\frac{45}{49}$ ...(3)

On solving (2) and (3), we get

$A=-\frac{45}{98},C=-\frac{15}{49}$

$\therefore \frac{(x-7)(x-1)}{{(x-5)}^2{(x+2)}^2}=-\frac{45}{98}.\frac{1}{x-5}-\frac{8}{49}.\frac{1}{{(x-5)}^2}-\frac{15}{49}.\frac{1}{x+2}+\frac{27}{49}.\frac{1}{{(x+2)}^2}$

Integrating both sides, we get

$\int \frac{(x-7)(x-1)}{{(x-5)}^2{(x+2)}^2}dx=-\frac{45}{98}\int \frac{1}{x-5}dx-\frac{8}{49}\int \frac{1}{{(x-5)}^2}dx-\frac{15}{49}\int \frac{1}{x+2}dx+\frac{27}{49}\int \frac{1}{{(x+2)}^2}dx$

$=-\frac{45}{98}\log |x-5|+\frac{8}{49}.\frac{1}{x-5}-\frac{15}{49}\log |x+2|-\frac{27}{49}.\frac{1}{x+2}+c$

$\therefore P=-\frac{45}{98},Q=\frac{8}{49},R=\frac{15}{49}$ and $S=\frac{27}{49}$