JEE Main 2015 — Mathematics Calculus

$\displaystyle f(x)= \int _{1}^{x}\frac{\mathrm{log}⁡t}{1+t} dt$f(x)=∫1x​1+tlog⁡t​dt $\displaystyle \Rightarrow f(x)=\int _{1}^{x}\frac{\mathrm{log}⁡z}{(1+z)}dz$⇒f(x)=∫1x​(1+z)log⁡z​dz

And $\displaystyle f(\frac{1}{x})= \int _{1}^{\frac{1}{x}}\frac{\mathrm{log}⁡t}{1+t} dt$f(x1​)=∫1x1​​1+tlog⁡t​dt

Put $t=\frac{1}{z}$t=z1​

$$dt= -\frac{1}{{z}^{2}} dz$$dt=−z21​dz

$$f(x)= \int _{1}^{x}\frac{\mathrm{log}⁡z}{{z}^{2}(1+ \frac{1}{z})} \cdot dz$$f(x)=∫1x​z2(1+z1​)log⁡z​⋅dz

$$f(x)= \int _{1}^{x}\frac{\mathrm{log}⁡z}{z(1+z)} dz$$f(x)=∫1x​z(1+z)log⁡z​dz

$$f(x)+f(\frac{1}{x})= \int _{1}^{x}\mathrm{log}⁡z [\frac{1}{1+z}+\frac{1}{z(1+z)}] dz$$f(x)+f(x1​)=∫1x​log⁡z[1+z1​+z(1+z)1​]dz

$$= \int _{1}^{x}\frac{1}{z}\mathrm{log}⁡z dz$$=∫1x​z1​log⁡zdz

Put $\mathrm{log}⁡z=P$log⁡z=P$\Rightarrow \frac{1}{z} dz=dP$⇒z1​dz=dP

$$\int _{1}^{x}P\cdot dP={(\frac{{P}^{2}}{2})}_{1}^{x}=\frac{1}{2} {({\mathrm{log}}^{2}⁡z)}_{1}^{x}= \frac{{(\mathrm{log}⁡x)}^{2}}{2}$$∫1x​P⋅dP=(2P2​)1x​=21​(log2⁡z)1x​=2(log⁡x)2​