Q112 / 160 · AP EAMCET 2024 (21 May Shift 1) · Mathematics
If $729 \int_1^3 \frac{1}{x^3\left(x^2+9\right)^2} d x=a+\log b$, then $a-b=$
- Correct answerA4
- B$-\frac{4}{5}$
- C$\frac{4}{5}$
- D-4
Solution
$\mathrm{I}=729 \int_1^3 \frac{1}{x^3\left(x^2+9\right)^2} d x$
Let $\frac{1}{x^3\left(x^2+9\right)^2}=\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{x^2}+\frac{C}{x^3}+\frac{\mathrm{D} x+\mathrm{E}}{x^2+9}+\frac{\mathrm{F} x+\mathrm{G}}{\left(x^2+9\right)^2}$ $\begin{aligned} & \Rightarrow 1=x^3\left(x^2+9\right)(\mathrm{D} x+\mathrm{E})+x^3(\mathrm{~F} x+\mathrm{G}) \\ & \qquad+x^2\left(x^2+9\right)^2 \mathrm{~A}+x\left(x^2+9\right)^2 \mathrm{~B}+\left(x^2+9\right)^2 \mathrm{C} \\ & \Rightarrow 1=x^6(\mathrm{~A}+\mathrm{D})+x^5(\mathrm{~B}+\mathrm{E})+x^4(18 \mathrm{~A}+\mathrm{C}+9 \mathrm{D}+\mathrm{F}) \\ & +x^3(18 \mathrm{~B}+9 \mathrm{E}+\mathrm{G})+x^2(81 \mathrm{~A}+18 \mathrm{C})+81 x \mathrm{~B}+81 \mathrm{C} \end{aligned}$
On comparing, $\begin{aligned} & \mathrm{A}+\mathrm{D}=0, \mathrm{~B}+\mathrm{E}=0 ; 18 \mathrm{~A}+\mathrm{C}+9 \mathrm{D}+\mathrm{F}=0 \\ & 18 \mathrm{~B}+9 \mathrm{E}+\mathrm{G}=0,81 \mathrm{~A}+18 \mathrm{C}=0 \\ & 81 \mathrm{~B}=0,81 \mathrm{C}=1 \\ & \Rightarrow \mathrm{~A}=\frac{-2}{729}, B=0, C=\frac{1}{81}, D=\frac{2}{729}, E=0, F=\frac{1}{81} \\ & \mathrm{G}=0 \\ & \therefore \mathrm{I}=\int_1^3 \frac{-2}{x} d x+9 \int_1^3 \frac{1}{x^3} d x+2 \int_1^3 \frac{x}{x^2+9} d x+9 \int_1^3 \frac{x}{\left(x^2+9\right)^2} d x \end{aligned}$ $\begin{aligned} \mathrm{I} & =\left[\log \frac{1}{x^2} \frac{-9}{2} \frac{1}{x^2}+\log \left(x^2+9\right) \frac{-9}{2\left(x^2+9\right)}\right]_1^3 \\ \mathrm{I} & =\left[\log \left(\frac{x^2+9}{x^2}\right)-\frac{9}{2} \frac{1}{x^2}-\frac{9}{2} \frac{1}{x^2+9}\right]_1^3 \\ \mathrm{I} & =\left[\log 2-\frac{1}{2}-\frac{1}{4}-\log 10+\frac{9}{2}+\frac{9}{20}\right]=\left[\log \frac{1}{5}+\frac{21}{5}\right] \\ & \Rightarrow a=\frac{21}{5}, b=\frac{1}{5} \Rightarrow a-b=\frac{20}{5}=4\end{aligned}$
Let $\frac{1}{x^3\left(x^2+9\right)^2}=\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{x^2}+\frac{C}{x^3}+\frac{\mathrm{D} x+\mathrm{E}}{x^2+9}+\frac{\mathrm{F} x+\mathrm{G}}{\left(x^2+9\right)^2}$ $\begin{aligned} & \Rightarrow 1=x^3\left(x^2+9\right)(\mathrm{D} x+\mathrm{E})+x^3(\mathrm{~F} x+\mathrm{G}) \\ & \qquad+x^2\left(x^2+9\right)^2 \mathrm{~A}+x\left(x^2+9\right)^2 \mathrm{~B}+\left(x^2+9\right)^2 \mathrm{C} \\ & \Rightarrow 1=x^6(\mathrm{~A}+\mathrm{D})+x^5(\mathrm{~B}+\mathrm{E})+x^4(18 \mathrm{~A}+\mathrm{C}+9 \mathrm{D}+\mathrm{F}) \\ & +x^3(18 \mathrm{~B}+9 \mathrm{E}+\mathrm{G})+x^2(81 \mathrm{~A}+18 \mathrm{C})+81 x \mathrm{~B}+81 \mathrm{C} \end{aligned}$
On comparing, $\begin{aligned} & \mathrm{A}+\mathrm{D}=0, \mathrm{~B}+\mathrm{E}=0 ; 18 \mathrm{~A}+\mathrm{C}+9 \mathrm{D}+\mathrm{F}=0 \\ & 18 \mathrm{~B}+9 \mathrm{E}+\mathrm{G}=0,81 \mathrm{~A}+18 \mathrm{C}=0 \\ & 81 \mathrm{~B}=0,81 \mathrm{C}=1 \\ & \Rightarrow \mathrm{~A}=\frac{-2}{729}, B=0, C=\frac{1}{81}, D=\frac{2}{729}, E=0, F=\frac{1}{81} \\ & \mathrm{G}=0 \\ & \therefore \mathrm{I}=\int_1^3 \frac{-2}{x} d x+9 \int_1^3 \frac{1}{x^3} d x+2 \int_1^3 \frac{x}{x^2+9} d x+9 \int_1^3 \frac{x}{\left(x^2+9\right)^2} d x \end{aligned}$ $\begin{aligned} \mathrm{I} & =\left[\log \frac{1}{x^2} \frac{-9}{2} \frac{1}{x^2}+\log \left(x^2+9\right) \frac{-9}{2\left(x^2+9\right)}\right]_1^3 \\ \mathrm{I} & =\left[\log \left(\frac{x^2+9}{x^2}\right)-\frac{9}{2} \frac{1}{x^2}-\frac{9}{2} \frac{1}{x^2+9}\right]_1^3 \\ \mathrm{I} & =\left[\log 2-\frac{1}{2}-\frac{1}{4}-\log 10+\frac{9}{2}+\frac{9}{20}\right]=\left[\log \frac{1}{5}+\frac{21}{5}\right] \\ & \Rightarrow a=\frac{21}{5}, b=\frac{1}{5} \Rightarrow a-b=\frac{20}{5}=4\end{aligned}$
Hence, the correct answer is option: 'A'.