If $\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}$ $=A\left(\tan ^{-1}\left(\frac{x-1}{3}\right)+\frac{f(x)}{x^{2}-2 x+10}\right)+C$ Where $C$ is a constant of integration, then:

(C) $A=\frac{1}{54}$ and $f(x)=3(x-1)$

(A) $\mathrm{A}=\frac{1}{27}$ and $\mathrm{f}(\mathrm{x})=-(\mathrm{x}-1)$

(B) $A=\frac{1}{54}$ and $f(x)=9(x-1)^{2}$

(D) $A=\frac{1}{81}$ and $f(x)=3(x-1)$

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Indefinite Integrals

\int \frac{d x}{( x ^{2} - 2 x + 10 ) ^{2}}

= A (tan^{- 1} (\frac{x - 1}{3}) + \frac{f ( x )}{x ^{2} - 2 x + 10}) + C

Where

C

is a constant of integration, then:

Answer: (C)

Complete the square:

\int \frac{d x}{( x ^{2} - 2 x + 10 ) ^{2}} = \int \frac{d x}{(( x - 1 ) ^{2} + 9 ) ^{2}}

. Substituting

x - 1 = 3 tan θ

d x = 3 sec^{2} θ d θ

, the integral becomes

\int \frac{3 s e c ^{2} θ}{81 s e c ^{4} θ} d θ = \frac{1}{27} \int cos^{2} θ d θ = \frac{1}{54} (θ + \frac{s i n 2 θ}{2}) + C

. Back-substituting with

θ = tan^{- 1} \frac{x - 1}{3}

and

sin 2 θ = \frac{6 ( x - 1 )}{x ^{2} - 2 x + 10}

gives

\frac{1}{54} (tan^{- 1} \frac{x - 1}{3} + \frac{3 ( x - 1 )}{x ^{2} - 2 x + 10}) + C

, so

A = \frac{1}{54}

and

f (x) = 3 (x - 1)

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option