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For x>0,let  f(x)=\int_{1}^{x}\frac{log\, t}{1+t}dt.\: Then \: f(x)+f\left ( \frac{1}{x} \right ) is \: equal to : 

  • Option 1)

    \frac{1}{4}\left ( log\, x \right )^{2}

  • Option 2)

    \frac{1}{2}\left ( log\, x \right )^{2}

  • Option 3)

    log\, x \right

  • Option 4)

    \frac{1}{4}\, log\, x^{2}

 

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 )

 

 

 f(x)=\int_{1}^{x}\frac{logt}{1+t}dx

f(\frac{1}{x})=\int_{1}^{\frac{1}{x}}\frac{logt}{1+t}dt

Put t=\frac{1}{z} (substitution)

=\int_{1}^{z}\left ( \frac{-log \ z}{1+\frac{1}{z}} \right )\times \frac{-1}{z^{2}}dz

=\int_{1}^{z}\left ( \frac{+log \ z}{1+z} \right )\times \frac{1}{z}dz

f(x)+f(\frac{1}{x})=\int_{1}^{x}\frac{logt}{1+t}\left ( 1+\frac{1}{t} \right )dt

=\int_{1}^{x}\frac{logt}{t} dt

=\frac{(logx)^{2}}{2} -0 


Option 1)

\frac{1}{4}\left ( log\, x \right )^{2}

This is incorrect option

Option 2)

\frac{1}{2}\left ( log\, x \right )^{2}

This is correct option

Option 3)

log\, x \right

This is incorrect option

Option 4)

\frac{1}{4}\, log\, x^{2}

This is incorrect option

Posted by

divya.saini

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