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Indefinite Integration

Section: Mathematics

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Question : 25 of 89

Marks: +1, -0

If

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

\sqrt{3 A^2+4 D^2+5 C^2+B^2}=

[AP EAPCET 22 May 2025 S1]

$\;\frac{3}{2}$
$\;\frac{1}{2}$
$1$
$2$

Solution:

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

\;x+1=A(x-1)(x^2+1)+B(x^2+1)+(C x+D)(x-1)^2 ..... (i)

Put

x=1

2=2B \Rightarrow B=1

In Eq. (i) put

x=0

\;1=-A+B+D ...(ii)

\Rightarrow \;\; A=D ...(ii)

In Eq. (i) put

x=-1

0\;=-4A+2-4C+4D

4C\;=2 \Rightarrow C=\frac{1}{2} ...(iii)

In Eq. (i) put

x=2

3=5A+5+1+D \Rightarrow 6A=-3

A=-\;\frac{1}{2}=D

\;\sqrt{3 A^2+4 D^2+5 C^2+B^2}

\;=\sqrt{\;\frac{3}{4}+\;\frac{4}{4}+\;\frac{5}{4}+1}

\;=\sqrt{\;\frac{16}{4}}=2

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