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Let    F(x)=f(x)+f\left ( \frac{1}{x} \right ),where\; f(x)=\int_{1}^{x}\frac{log\, t}{1+t}dt,   Then F(e) equals

  • Option 1)

    1\;

  • Option 2)

    \; 2\;

  • Option 3)

    \; 1/2\;

  • Option 4)

    \; 0

 

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)=f(x)+f\left (\frac{1}{x} \right )

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

Put t=\frac{1}{z},f\left (\frac{1}{x} \right )=\int_{1}^{x}-\frac{logz}{1+\frac{1}{z}}\times \frac{-1}{z^{2}}d{z}

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

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

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

=\frac{(logx)^{2}}{2}-0

Put x=e;we get =\frac{(loge)^{2}}{2}=\frac{1}{2}


Option 1)

1\;

This option is incorrect

Option 2)

\; 2\;

This option is incorrect

Option 3)

\; 1/2\;

This option is correct

Option 4)

\; 0

This option is incorrect

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prateek

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