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Indefinite Integration

Section: Mathematics

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Question : 12 of 89

Marks: +1, -0

If

\frac{3x+1}{(x-1)(x^2+2)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+2}

5(A-B)=

[AP EAPCET 23 May 2025 S2]

$A + C$
$8C$
$C + 8$
$\frac{C}{8}$

Solution:

\because \frac{3x+1}{(x-1)(x^2+2)} = \frac{A}{x-1} + \frac{Bx+c}{x^2+2}

\Rightarrow 3x+1 = A(x^2+2) + (Bx+C)(x-1)

\Rightarrow 3x+1 = (A+B)x^2 + (C-B)x + (2A-C)

Comparing both sides, we get

A+B = 0

\Rightarrow A = -B \cdot\cdot\cdot\cdot\cdot\cdot\cdot (i)

and

C-B = 3 \cdot\cdot\cdot\cdot\cdot\cdot\cdot (ii)

and

2A-C = 1 \cdot\cdot\cdot\cdot\cdot\cdot\cdot (iii)

put the value of

A

from Eq. (i) to Eq. (iii)

\Rightarrow -2B-C = 1

\Rightarrow C = -2B-1 \cdot\cdot\cdot\cdot\cdot\cdot\cdot (iv)

By Eqs. (ii) and (iv),

We get

(-2B-1)-B = 3 \Rightarrow B = -\frac{4}{3}

and

A = -\left(-\frac{4}{3}\right) = \frac{4}{3}

Also,

C = -2B-1 = -2\left(-\frac{4}{3}\right)-1

= \frac{8}{3} - 1 = \frac{5}{3}

thus,

5(A-B) = 5\left(\frac{4}{3} - \left(-\frac{4}{3}\right)\right)

= 5\left(\frac{8}{3}\right) = \frac{40}{3}

and

8C = 8 \times \frac{5}{3} = \frac{40}{3}

Therefore,

5(A-B) = 8C

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