JEE Main 2016 — Mathematics Calculus

From the given equation

$$2\int _{0}^{1}{\mathrm{tan}}^{-1}⁡xdx= \int _{0}^{1}(\frac{\pi }{2}-{\mathrm{tan}}^{-1}⁡(1-x+{x}^{2}))dx$$2∫01​tan−1⁡xdx=∫01​(2π​−tan−1⁡(1−x+x2))dx

$$\Rightarrow 2\int _{0}^{1}{\mathrm{tan}}^{-1}⁡xdx= \int _{0}^{1}\frac{\pi }{2} dx- \int _{0}^{1}{\mathrm{tan}}^{-1}⁡(1-x+{x}^{2})dx$$⇒2∫01​tan−1⁡xdx=∫01​2π​dx−∫01​tan−1⁡(1−x+x2)dx

$$\Rightarrow \int _{0}^{1}{\mathrm{tan}}^{-1}⁡(1-x+{x}^{2})dx=\frac{\pi }{2}-2 \int _{0}^{1}{\mathrm{tan}}^{-1}⁡xdx\ldots (i)$$⇒∫01​tan−1⁡(1−x+x2)dx=2π​−2∫01​tan−1⁡xdx…(i)

Let $\displaystyle {I}_{1}= \int _{0}^{1}{\mathrm{tan}}^{-1}⁡xdx$I1​=∫01​tan−1⁡xdx

$$={[({\mathrm{tan}}^{-1}⁡x)x]}_{0}^{1}-\int _{0}^{1}\frac{1}{1+{x}^{2}}xdx$$=[(tan−1⁡x)x]01​−∫01​1+x21​xdx

$$=\frac{\pi }{4}- \int _{0}^{1}\frac{x}{1+{x}^{2}}dx$$=4π​−∫01​1+x2x​dx

$$=\frac{\pi }{4}-\frac{1}{2} \int _{0}^{1}\frac{2x}{1+{x}^{2}}dx$$=4π​−21​∫01​1+x22x​dx

$$=\frac{\pi }{4}-\frac{1}{2} {[\mathrm{ln}(1+{x}^{2})]}_{0}^{1}$$=4π​−21​[ln(1+x2)]01​

$$=\frac{\pi }{4}-\frac{1}{2}\mathrm{ln}⁡2$$=4π​−21​ln⁡2

By equation $(i)$(i),

$\displaystyle {\int }_{0}^{1}{\mathrm{tan}}^{-1}⁡(1-x+{x}^{2})dx=$∫01​tan−1⁡(1−x+x2)dx= $\frac{\pi }{2}-2[\frac{\pi }{4}-\frac{1}{2}\mathrm{ln}2]=\mathrm{ln}⁡2$2π​−2[4π​−21​ln2]=ln⁡2