JEE Main 2020 — Mathematics Calculus

$$I=\int {\mathrm{sin}}^{-1}(\frac{\sqrt{x}}{1+x})dx$$I=∫sin−1(1+xx​​)dx

$$=\int \underset{I}{\underset{⏟}{{\mathrm{tan}}^{-1}(\sqrt{x})}}\underset{II}{\underset{⏟}{(1)}}dx$$=∫I⏟tan−1(x​)​​II⏟(1)​​dx

$$=x{\mathrm{tan}}^{-1}\sqrt{x}-\int \frac{1}{1+x}.\frac{1}{2\sqrt{x}}.xdx+C$$=xtan−1x​−∫1+x1​.2x​1​.xdx+C

$\displaystyle =x{\mathrm{tan}}^{-1}\sqrt{x}-\frac{1}{2}\int \frac{t.2t.dt}{1+{t}^{2}}+C$=xtan−1x​−21​∫1+t2t.2t.dt​+C $($(putting $x={t}^{2}\Rightarrow dx=2tdt)$x=t2⇒dx=2tdt)

$$=x{\mathrm{tan}}^{-1}\sqrt{x}-\int \frac{{t}^{2}}{1+{t}^{2}}dt+C$$=xtan−1x​−∫1+t2t2​dt+C

$$=x{\mathrm{tan}}^{-1}\sqrt{x}-t+{\mathrm{tan}}^{-1}t+C$$=xtan−1x​−t+tan−1t+C

$$=x{\mathrm{tan}}^{-1}\sqrt{x}-\sqrt{x}+{\mathrm{tan}}^{-1}\sqrt{x}+C$$=xtan−1x​−x​+tan−1x​+C

$$=(x+1){\mathrm{tan}}^{-1}\sqrt{x}-\sqrt{x}+C$$=(x+1)tan−1x​−x​+C

Comparing with $A(x){\mathrm{tan}}^{-1}(\sqrt{x})+B(x)+C$A(x)tan−1(x​)+B(x)+C, we get

$$(A(x),B(x))=(x+1,-\sqrt{x})$$(A(x),B(x))=(x+1,−x​)