Question

༒ Bʀᴀɪɴʟʏ Sᴛᴀʀs

༒Mᴏᴅʀᴀᴛᴇʀs




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If
$\large\bf \int \frac{dx}{( {x}^{2} - 2x + 10) ^{2} } \\ \\ = \bf A \bigg \{{tan}^{ - 1} \bigg( \frac{x + 1}{3} \bigg) + \frac{f(x)}{ {x}^{2} - 2x + 10} \bigg \} + C$
Then ➽

$\bf a)A = \frac{1}{27} \: and \: f(x) = 9(x - 1) \\ \\ \bf b)A = \frac{1}{81} \: and \: f(x) = 3(x + 1) \\ \\ \bf c)A = \frac{1}{54} \: and \: f(x) = 3(x - 1) \\ \\ \bf d)A = \frac{1}{54} \: and \: f(x) = 9(x - 1) {}^{2}$
​

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

$\displaystyle \sf{\int\limits_{}^{} \: \frac{dx}{ {( {x}^{2} - 2x + 10) }^{2} } = A \bigg \{ { \tan}^{ - 1} \bigg( \frac{x - 1}{3} \bigg) + \frac{f(x)}{ {x}^{2} - 2x + 10} \bigg \}+ C}$

$\displaystyle \sf{a) \: \:A = \frac{1}{27} \: \: \: and \: \: \: f(x) = 9(x - 1) }$

$\displaystyle \sf{b) \: \:A = \frac{1}{81} \: \: \: and \: \: \: f(x) = 3(x + 1) }$

$\displaystyle \sf{c) \: \:A = \frac{1}{54} \: \: \: and \: \: \: f(x) = 3(x - 1) }$

$\displaystyle \sf{d) \: \:A = \frac{1}{54} \: \: \: and \: \: \: f(x) = 9{(x - 1) }^{2} }$

EVALUATION

Here the given expression is

$\displaystyle \sf{\int\limits_{}^{} \: \frac{dx}{ {( {x}^{2} - 2x + 10) }^{2} } = A \bigg \{ { \tan}^{ - 1} \bigg( \frac{x - 1}{3} \bigg) + \frac{f(x)}{ {x}^{2} - 2x + 10} \bigg \}+ C} \: \: \: - - - (1)$

First we integrate the LHS of the above expression

$\displaystyle \sf{\int\limits_{}^{} \: \frac{dx}{ {( {x}^{2} - 2x + 10) }^{2} } }$

$\displaystyle \sf{ = \int\limits_{}^{} \: \frac{dx}{ {( {x}^{2} - 2x + 1 + 9) }^{2} }}$

$\displaystyle \sf{ = \int\limits_{}^{} \frac{dx}{ { \big[ {(x - 1)}^{2} + {3}^{2} \big] }^{2} } }$

Let x - 1 = 3 tan θ

⇒ dx = 3 sec² θ dθ

Thus above becomes

$\displaystyle \sf{\int\limits_{}^{} \: \frac{dx}{ {( {x}^{2} - 2x + 10) }^{2} } }$

$\displaystyle \sf{ = \int\limits_{}^{} \frac{dx}{ { \big[ {(x - 1)}^{2} + {3}^{2} \big] }^{2} } }$

$\displaystyle \sf{ = \int\limits_{}^{} \frac{3 { \sec}^{2} \theta \: d \theta }{ { \big[ {(3 \tan \theta)}^{2} + {3}^{2} \big] }^{2} } }$

$\displaystyle \sf{ = \int\limits_{}^{} \frac{3 { \sec}^{2} \theta \: d \theta }{ { 81(1 + { \tan }^{2} \theta ) }^{2} } }$

$\displaystyle \sf{ = \int\limits_{}^{} \frac{ { \sec}^{2} \theta \: d \theta }{ { 27( { \sec }^{2} \theta ) }^{2} } }$

$\displaystyle \sf{ = \frac{1}{27} \int\limits_{}^{} \frac{ \: d \theta }{ { \sec }^{2} \theta } }$

$\displaystyle \sf{ = \frac{1}{27} \int\limits_{}^{} \: { \cos }^{2} \theta \: \: d \theta }$

$\displaystyle \sf{ = \frac{1}{54} \int\limits_{}^{} \: 2{ \cos }^{2} \theta \: \: d \theta }$

$\displaystyle \sf{ = \frac{1}{54} \int\limits_{}^{} \: (1 + \cos 2 \theta) \: \: d \theta }$

$\displaystyle \sf{ = \frac{1}{54} \bigg(\theta + \frac{ \sin 2\theta}{2} \bigg) +C \: \: }$

$\displaystyle \sf{ = \frac{1}{54} \bigg(\theta + \frac{ 2\sin \theta \cos \theta}{2} \bigg) +C \: \: }$

$\displaystyle \sf{ = \frac{1}{54} \bigg(\theta + \sin \theta \cos \theta \bigg) +C \: \: }$

Now x - 1 = 3 tan θ gives

$\displaystyle \sf{ \theta = { \tan}^{ - 1} \bigg( \frac{x - 1}{3} \bigg) \: \:, \: \sin \theta = \frac{x - 1}{ \sqrt{ {x}^{2} - 2x + 10} } \:, \: \cos \theta = \frac{3}{ \sqrt{ {x}^{2} - 2x + 10 } } }$

Thus we get

$\displaystyle \sf{\int\limits_{}^{} \: \frac{dx}{ {( {x}^{2} - 2x + 10) }^{2} } }$

$\displaystyle \sf{ = \frac{1}{54} \bigg(\theta + \sin \theta \cos \theta \bigg) +C \: \: }$

$\displaystyle \sf{ = \frac{1}{54} \bigg \{ { \tan}^{ - 1} \bigg( \frac{x - 1}{3} \bigg) + \frac{3(x - 1)}{ {x}^{2} - 2x + 10} \bigg \}+ C} \: \: \:$

Thus we get

$\boxed{ \: \: \displaystyle \sf{\int\limits_{}^{} \: \frac{dx}{ {( {x}^{2} - 2x + 10) }^{2} } = \frac{1}{54} \bigg \{ { \tan}^{ - 1} \bigg( \frac{x - 1}{3} \bigg) + \frac{3(x - 1)}{ {x}^{2} - 2x + 10} \bigg \}+ C} \: \: }$

Comparing with Equation 1 we get

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

FINAL ANSWER

Hence the correct option is

$\displaystyle \sf{c) \: \:A = \frac{1}{54} \: \: \: and \: \: \: f(x) = 3(x - 1) }$

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Answer 2

$\large\underline{\sf{Solution-}}$

Appropriate Question

Given that,

$\large\bf \int \frac{dx}{( {x}^{2} - 2x + 10) ^{2} } \\ \\ = \bf A \bigg \{{tan}^{ - 1} \bigg( \frac{x - 1}{3} \bigg) + \frac{f(x)}{ {x}^{2} - 2x + 10} \bigg \} + C$

Consider,

$\rm :\longmapsto\:\displaystyle\int\rm \dfrac{1}{ {( {x}^{2} - 2x + 10)}^{2} } \: dx$

can be rewritten as

$\rm \: = \:\displaystyle\int\rm \dfrac{1}{ {( {x}^{2} - 2x + 1 + 9)}^{2} } \: dx$

$\rm \: = \:\displaystyle\int\rm \dfrac{1}{ { \bigg( {(x - 1)}^{2} + 9\bigg)}^{2} } \: dx$

can be further rewritten as

$\rm \: = \:\displaystyle\int\rm \dfrac{1}{ { \bigg( {(x - 1)}^{2} + {3}^{2} \bigg)}^{2} } \: dx$

Now, we use method of Substitution, to evaluate this integral

$\rm :\longmapsto\:Put \: x - 1 = 3tany$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:dx = 3 {sec}^{2}y \: dy$

Now substituting the values, we have

$\rm \: = \: 3 \: \displaystyle\int\rm \dfrac{ {sec}^{2} y}{ {\bigg( {(3tany)}^{2} + 9\bigg)}^{2} } \: dy$

$\rm \: = \: 3\: \displaystyle\int\rm \dfrac{ {sec}^{2} y}{ {\bigg(9 {tan}^{2}y + 9\bigg)}^{2} } \: dy$

$\rm \: = \: \: \dfrac{3}{81} \displaystyle\int\rm \dfrac{ {sec}^{2} y}{ {\bigg({tan}^{2}y +1 \bigg)}^{2} } \: dy$

$\rm \: = \: \: \dfrac{1}{27} \displaystyle\int\rm \dfrac{ {sec}^{2} y}{ {\bigg({sec}^{2}y\bigg)}^{2} } \: dy$

$\rm \: = \: \: \dfrac{1}{27} \displaystyle\int\rm \dfrac{ 1}{ {\bigg({sec}^{2}y\bigg)} } \: dy$

$\rm \: = \: \: \dfrac{1}{27} \displaystyle\int\rm {cos}^{2}y \: dy$

$\rm \: = \: \: \dfrac{1}{54} \displaystyle\int\rm {2cos}^{2}y \: dy$

$\rm \: = \: \: \dfrac{1}{54} \displaystyle\int\rm (1 + cos2y) \: dy$

$\rm \: = \: \: \dfrac{1}{54}\bigg(y + \dfrac{sin2y}{2}\bigg) + c - - - (1)$

Now,

As

We assume that

$\rm :\longmapsto \: \: x - 1 = 3tany$

$\rm :\longmapsto\:tany = \dfrac{x - 1}{3}$

$\rm :\implies\:y = {tan}^{ - 1}\bigg(\dfrac{x - 1}{3} \bigg) - - - (2)$

and

We know

$\rm :\longmapsto\:sin2y = \dfrac{2tany}{1 + {tan}^{2} y}$

$\rm :\longmapsto\:sin2y = \dfrac{2 \times \dfrac{x - 1}{3} }{1 + {( \dfrac{x - 1}{3}) }^{2} }$

$\rm :\longmapsto\:sin2y = \dfrac{\dfrac{2x - 2}{3} }{1 + {\dfrac{ {x}^{2} + 1 - 2x}{9} }}$

$\rm :\longmapsto\:sin2y = \dfrac{\dfrac{2x - 2}{3} }{ {\dfrac{ 9 + {x}^{2} + 1 - 2x}{9} }}$

$\rm :\longmapsto\:sin2y = \dfrac{6(x - 1)}{ {x}^{2} - 2x + 10} - - - (3)$

On substituting the values from equation (2) and (3) in (1),

$\rm \: = \: \: \dfrac{1}{54}\bigg( {tan}^{ - 1}\dfrac{x - 1}{3} + \dfrac{3(x - 1)}{ {x}^{2} - 2x + 10} \bigg) + c$

Hence,

$\rm :\longmapsto\:\displaystyle\int\bf \dfrac{1}{ {( {x}^{2} - 2x + 10)}^{2} } \: dx$

$\bf \: = \: \: \dfrac{1}{54}\bigg( {tan}^{ - 1}\dfrac{x - 1}{3} + \dfrac{3(x - 1)}{ {x}^{2} - 2x + 10} \bigg) + c$

As it is given that

$\large\bf \int \frac{dx}{( {x}^{2} - 2x + 10) ^{2} } \\ \\ = \bf A \bigg \{{tan}^{ - 1} \bigg( \frac{x - 1}{3} \bigg) + \frac{f(x)}{ {x}^{2} - 2x + 10} \bigg \} + C$

So, on comparing, we get

$\: \: \: \: \: \: \red{ \boxed{ \bf{ \: A = \frac{1}{54} \: \: and \: \: f(x) = 3(x - 1)}}}$

Thus,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \underbrace{ \boxed{ \bf{ \: Option \: (c) \: is \: correct}}}}$