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\small \int_{1}^{2}\frac{dx}{\left ( x^{2}-2x+4 \right )^{\frac{3}{2}}} =\frac{k}{k+5},then k is equal to:

  • Option 1)

    1

  • Option 2)

    2

  • Option 3)

    3

  • Option 4)

    4

 

Answers (1)

best_answer

As learnt in concept

Integration by substitution -

The functions when on substitution of the variable of integration to some quantity gives any one of standard formulas.

 

 

- wherein

Since \int f(x)dx=\int f(t)dt=\int f(\theta )d\theta all variables must be converted into single variable ,\left ( t\, or\ \theta \right )

 

 

 \int_{1}^{2}\frac{dx}{\left ( x^{2}-2x+4 \right )\frac{3}{2}}

put t = x-1

=>\int_{0}^{1}\frac{dt}{\left ( \left ( t+1 \right )^{2}-2\left ( t+1 \right )+4 \right )^{\frac{3}{2}}}

=> \int_{0}^{1} \frac{dt}{\left ( t^{2}+3 \right )^{\frac{3}{2}}}

t=\sqrt{3}\tan \Theta

dt=\sqrt{3}\sec ^{2}\Theta \:d\Theta

=>\int_{0}^{\tan ^{-1}\frac{1}{\sqrt{3}}} \frac{\sqrt{3}\sec^{2}\theta \:d\theta }{3\sqrt{3}\sec ^{3}\theta }

=>\frac{1}{3}\int_{0}^{\tan ^{-1}\frac{1}{\sqrt{3}}} \cos \theta \:d\theta

=\frac{1}{3}\left [ \sin \theta \right ]^{\tan^{-1}\frac{1}{\sqrt{3}} }_{0}

=\frac{1}{3}\times \frac{1}{2}

=\frac{1}{6}

=\frac{K}{K+5}

=> K=1


Option 1)

1

Correct option

Option 2)

2

Incorrect option    

Option 3)

3

Incorrect option    

Option 4)

4

Incorrect option    

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