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Maths-

General

Easy

Question

Let $f left parenthesis x right parenthesis equals not stretchy integral fraction numerator x to the power of 2 end exponent d x over denominator left parenthesis 1 plus x to the power of 2 end exponent right parenthesis open parentheses 1 plus square root of 1 plus x to the power of 2 end exponent end root close parentheses end fraction$ and $f left parenthesis 0 right parenthesis equals 0$ , then the value of $f left parenthesis 1 right parenthesis$ be

$\log (1 + \sqrt{2})$
$\log (1 + \sqrt{2}) - \frac{π}{4}$
$\log (1 + \sqrt{2}) + \frac{π}{2}$
None of these

The correct answer is: $log invisible function application left parenthesis 1 plus square root of 2 right parenthesis minus fraction numerator pi over denominator 4 end fraction$

$f left parenthesis x right parenthesis equals not stretchy integral fraction numerator x to the power of 2 end exponent d x over denominator left parenthesis 1 plus x to the power of 2 end exponent right parenthesis open parentheses 1 plus square root of 1 plus x to the power of 2 end exponent end root close parentheses end fraction$
Let $x equals tan invisible function application theta comma d x equals sec to the power of 2 end exponent invisible function application theta d theta equals left parenthesis 1 plus x to the power of 2 end exponent right parenthesis. d theta$
$f left parenthesis x right parenthesis equals not stretchy integral fraction numerator x to the power of 2 end exponent d x over denominator left parenthesis 1 plus x to the power of 2 end exponent right parenthesis open parentheses 1 plus square root of 1 plus x to the power of 2 end exponent end root close parentheses end fraction equals not stretchy integral fraction numerator tan to the power of 2 end exponent invisible function application theta sec to the power of 2 end exponent invisible function application theta d theta over denominator sec to the power of 2 end exponent invisible function application theta left parenthesis 1 plus sec invisible function application theta right parenthesis end fraction$
$equals not stretchy integral fraction numerator tan to the power of 2 end exponent invisible function application theta d theta over denominator 1 plus sec invisible function application theta end fraction theta equals not stretchy integral fraction numerator sin to the power of 2 end exponent invisible function application theta d theta over denominator cos invisible function application theta left parenthesis 1 plus cos invisible function application theta right parenthesis end fraction equals not stretchy integral fraction numerator 1 minus cos to the power of 2 end exponent invisible function application theta d theta over denominator cos invisible function application theta left parenthesis 1 plus cos invisible function application theta right parenthesis end fraction$
$equals not stretchy integral fraction numerator left parenthesis 1 minus cos invisible function application theta right parenthesis d. theta over denominator cos invisible function application theta end fraction equals not stretchy integral sec invisible function application theta d theta minus not stretchy integral d theta$
$equals log invisible function application left parenthesis x plus square root of 1 plus x to the power of 2 end exponent end root right parenthesis minus tan to the power of negative 1 end exponent invisible function application x plus c$
$f left parenthesis 0 right parenthesis equals log invisible function application left parenthesis 0 plus square root of 1 plus 0 end root minus tan to the power of negative 1 end exponent invisible function application left parenthesis 0 right parenthesis plus c$
$0 equals log invisible function application 1 minus 0 plus c$ Þ $c equals 0$
$f left parenthesis 1 right parenthesis equals log invisible function application left parenthesis 1 plus square root of 1 plus 1 to the power of 2 end exponent end root right parenthesis minus tan to the power of negative 1 end exponent invisible function application left parenthesis 1 right parenthesis equals log invisible function application left parenthesis 1 plus square root of 2 right parenthesis minus fraction numerator pi over denominator 4 end fraction$ .

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