JEE Main 2015 · 10 Apr · Shift 1 · Mathematics Q75

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For $x>0$ , let $f(x)=\displaystyle\int_1^x\frac{\log t}{1+t}dt$ . Then $f(x)+f\left(\dfrac1x\right)$ is equal to :

\dfrac14(\log x)^2

\dfrac12(\log x)^2

\log x

\dfrac14\log x^2

Answer and worked solution

Official answer: B

$f(1/x)=\int_1^{1/x}\frac{\log t}{1+t}dt$ . With $t=\frac1u$ this becomes $\int_1^x\frac{\log u}{u(1+u)}du$ . Adding:
$f(x)+f(1/x)=\int_1^x\frac{\log u}u\,du=\frac12(\log x)^2.$

Topics:Properties of definite integrals Fundamental theorem of calculus

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JEE Main 2015 · 10 Apr · Shift 1, Mathematics question 75