Math, asked by SparklingBoy, 6 hours ago

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 \huge \mathfrak{  \red{Qu} \green{e} \pink{s} \blue{t} \purple{ion}}
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}

Answers

Answered by pulakmath007
55

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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amansharma264: Excellent
pulakmath007: Thank you Brother
Answered by mathdude500
32

\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}}}}


amansharma264: Great
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