$\begin{aligned} \text { Put } & x=\tan \theta \Rightarrow \mathrm{d} x=\sec ^2 \theta \mathrm{d} \theta \\ \therefore \quad \mathrm{f}(x) & =\int \frac{\tan ^2 \theta \sec ^2 \theta \mathrm{d} \theta}{\sec ^2 \theta(1+\sec \theta)} \\ & =\int \frac{\tan ^2 \theta \mathrm{d} \theta}{1+\sec \theta} \\ & =\int \frac{\sin ^2 \theta \mathrm{d} \theta}{\cos \theta(1+\cos \theta)} \\ & =\int \frac{1-\cos ^2 \theta \mathrm{d} \theta}{\cos \theta(1+\cos \theta)} \\ & =\int \frac{(1-\cos \theta) \mathrm{d} \theta}{\cos \theta} \\ & =\int \sec \theta \mathrm{d} \theta-\int \mathrm{d} \theta \\ & =\log |\sec \theta+\tan \theta|-\theta+\mathrm{c} \\ \mathrm{f}(x) & =\log \left|x+\sqrt{1+x^2}\right|-\tan ^{-1} x+\mathrm{c} \\ \therefore \quad \mathrm{f}(0) & =\log |0+\sqrt{1+0}|-\tan ^{-1}(0)+\mathrm{c} \\ \Rightarrow 0 & =\log 1-0+\mathrm{c} \Rightarrow \mathrm{c}=0 \\ \therefore \quad \mathrm{f}(x) & =\log \left|x+\sqrt{1+x^2}\right|-\tan ^{-1} x \\ \therefore \quad \mathrm{f}(1) & =\log \left|1+\sqrt{1+1^2}\right|-\tan ^{-1}(1) \\ \therefore \quad & \log (1+\sqrt{2})-\frac{\pi}{4}\end{aligned}$