JEE Main 2020 — Mathematics Calculus

$$f(x)=\int \frac{\sqrt{x}}{{(1+x)}^{2}}dx$$f(x)=∫(1+x)2x​​dx

Let $x={\mathrm{tan}}^{2}\theta$x=tan2θ

$$dx=2\mathrm{tan}\theta {\mathrm{sec}}^{2}\theta d\theta$$dx=2tanθsec2θdθ

$$f(x)=\int \frac{\mathrm{tan}\theta }{{(1+{\mathrm{tan}}^{2}\theta )}^{2}}.2\mathrm{tan}\theta {\mathrm{sec}}^{2}\theta d\theta$$f(x)=∫(1+tan2θ)2tanθ​.2tanθsec2θdθ

$$f(x)=\int \frac{\mathrm{tan}\theta }{{\mathrm{sec}}^{4}\theta }.2\mathrm{tan}\theta {\mathrm{sec}}^{2}\theta d\theta$$f(x)=∫sec4θtanθ​.2tanθsec2θdθ

$$f(x)=\int 2{\mathrm{tan}}^{2}\theta .{\mathrm{cos}}^{2}\theta d\theta$$f(x)=∫2tan2θ.cos2θdθ

$$f(x)=\int 2{\mathrm{sin}}^{2}\theta d\theta$$f(x)=∫2sin2θdθ

$$f(x)=\int (1-\mathrm{cos}2\theta )d\theta$$f(x)=∫(1−cos2θ)dθ

$$f(x)=\theta -\frac{\mathrm{sin}2\theta }{2}+C=\theta -\frac{\mathrm{tan}\theta }{1+{\mathrm{tan}}^{2}\theta }+C$$f(x)=θ−2sin2θ​+C=θ−1+tan2θtanθ​+C

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

Now, $f(3)-f(1)={\mathrm{tan}}^{-}\sqrt{3}-\frac{\sqrt{3}}{1+3}-{\mathrm{tan}}^{-1}\sqrt{1}+\frac{1}{1+1}$f(3)−f(1)=tan−3​−1+33​​−tan−11​+1+11​

$$=\frac{\pi }{3}-\frac{\pi }{4}+\frac{1}{2}-\frac{\sqrt{3}}{4}=\frac{\pi }{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}$$=3π​−4π​+21​−43​​=12π​+21​−43​​