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TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

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Question : 74 of 160

Marks: +1, -0

\int \frac{9x+15}{x^3-6x-9}dx=A\log|g(x)|+B\log|f(x)+C

\frac{(A-B)g(4)}{f(-1)}=

$3$
$\frac{1}{7}$
$1$
$\frac{3}{7}$

Solution:

We have,

\int \; \frac{9x+15}{x^{3}-6x-9} \; d x = A \log |g(x)| + B \log |f(x)| + C

Now,

\; \frac{9x+15}{x^{3}-6x-9} = \; \frac{9x+15}{(x-3)(x^{2}+3x+3)}

\therefore \; \frac{9x+15}{(x-3)(x^{2}+3x+3)} = \; \frac{A}{x-3} + \; \frac{Bx+C}{x^{2}+3x+3}

\Rightarrow 9x+15 = A(x^{2}+3x+3) + (Bx+c)(x-3)

\Rightarrow 9x+15 = x^{2}(A+B) + x(3A-3B+C) + (3A-3C)

On equating coefficients of

x^{2}, x

and constant term, we get

A+B=0, 3A-3B+C=9

and

3A-3C=15

On solving above three equations

We get,

A=2, B=-2

and

C=-3

\therefore \int \; \frac{9x+15}{(x-3)(x^{2}+3x+3)} d x

\;\; = \;\; \int \frac{2}{x-3} d x + \int \frac{-2x-3}{x^{2}+3x+3} d x

\;\; = \int \frac{2}{x-3} d x - \int \frac{2x+3}{x^{2}+3x+3} d x

\;\; = 2 \log |x-3| - \log |x^{2}+3x+3| + C

\therefore A \;\; =2, B=-1, g(x)=x-3 \;\text{ and }\; f(x)=x^{2}+3x+3

\therefore \;\; \; \frac{(A-B)g(4)}{f(-1)} = \frac{(2+1)(4-3)}{(-1)^{2}+3(-1)+3} = 3

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