JEE Mains · Maths · STD 12 - 7.2 definite integral
For \(x > 0\) , let \(f(x)\, = \,\int\limits_1^x {\frac{{\log \,t}}{{1 + t}}} \,dt\) then \(f(x)\, + \,f\left( {\frac{1}{x}} \right)\) is equal to
- A \(\frac{1}{4}\,{(\log \,x)^2}\)
- B \(\,\log \,x\)
- C \(\frac{1}{2}\,{(\log \,x)^2}\)
- D \(\frac{1}{4}\,\log \,{x^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}\,{(\log \,x)^2}\)
Step-by-step Solution
Detailed explanation
\(f\left( {\frac{1}{x}} \right) = \int\limits_1^{1/x} {\frac{{\ln t}}{{1 + t}}dt} \) \(\text { Let } t=\frac{1}{z}\) \(\mathrm{dt}=-\frac{1}{\mathrm{z}^{2}} \mathrm{d} \mathrm{z}\) \(f\left( x \right) = \int\limits_1^x {\frac{{\ln z}}{{z(z + 1)}}dz} \)…
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