JEE Main 2025 — Mathematics Calculus

$$\begin{aligned} & 4 \int_0^1 \frac{1}{\sqrt{3+\mathrm{x}^2}+\sqrt{1+\mathrm{x}^2}} \mathrm{dx}-3 \ln \sqrt{3} \\ & =4 \int_0^1 \frac{\sqrt{3+\mathrm{x}^2}-\sqrt{1+\mathrm{x}^2}}{\left(3+\mathrm{x}^2\right)-\left(1-\mathrm{x}^2\right)} \mathrm{dx}-\frac{3}{2} \ln 3 \\ & = \\ & 2\left[\left\{\frac{\mathrm{x}}{2} \sqrt{3+\mathrm{x}^2}+\frac{3}{2} \ln \left(\mathrm{x}+\sqrt{3+\mathrm{x}^2}\right)\right\}_0^1\right. \\ & \left.\quad-\left\{\frac{\mathrm{x}}{2} \sqrt{1+\mathrm{x}^2}+\frac{1}{2} \ln \left(\mathrm{x}+\sqrt{1+\mathrm{x}^2}\right)\right\}_0^1\right]-\frac{3}{2} \ln 3\end{aligned}$$​4∫01​3+x2​+1+x2​1​dx−3ln3​=4∫01​(3+x2)−(1−x2)3+x2​−1+x2​​dx−23​ln3=2[{2x​3+x2​+23​ln(x+3+x2​)}01​−{2x​1+x2​+21​ln(x+1+x2​)}01​]−23​ln3​

$$\begin{aligned}= & 2\left[\left\{\frac{1}{2} \sqrt{4}+\frac{3}{2} \ln (1+\sqrt{4})\right\}-\left\{0+\frac{3}{2} \ln \sqrt{3}\right\}\right. \\ & \left.-\left\{\frac{1}{2} \sqrt{2}+\frac{1}{2} \ln (1+\sqrt{2})\right\}+\left\{0+\frac{1}{2}(0)\right\}\right]-\frac{3}{2} \ln 3\end{aligned}$$=​2[{21​4​+23​ln(1+4​)}−{0+23​ln3​}−{21​2​+21​ln(1+2​)}+{0+21​(0)}]−23​ln3​

$$\begin{aligned} & =2\left[1+\frac{3}{2} \ln 3-\frac{3}{4} \ln 3-\frac{1}{\sqrt{2}}-\frac{1}{2} \ln (1+\sqrt{2})\right]-\frac{3}{2} \ln 3 \\ & =2+3 \ln 3-\frac{3}{2} \ln 3-\sqrt{2}-\ln (1+\sqrt{2})-\frac{3}{2} \ln 3 \\ & =2-\sqrt{2}-\ln (1+\sqrt{2})\end{aligned}$$​=2[1+23​ln3−43​ln3−2​1​−21​ln(1+2​)]−23​ln3=2+3ln3−23​ln3−2​−ln(1+2​)−23​ln3=2−2​−ln(1+2​)​