文字 d を含む積分は辛い

やってみよう

$ \int \sqrt{d^2+1}\mathrm d{d}

$ =d\sqrt{d^2+1}-\int{d}\cdot\frac{2d}{2\sqrt{d^2+1}}\mathrm d{d}

$ =d\sqrt{d^2+1}-\int\frac{d^2}{\sqrt{d^2+1}}\mathrm d{d}

$ =d\sqrt{d^2+1}-\int\frac{(d^2+1)-1}{\sqrt{d^2+1}}\mathrm d{d}

$ =d\sqrt{d^2+1}-\int\sqrt{d^2+1}\mathrm d{d}+\int\frac{1}{\sqrt{d^2+1}}\mathrm d{d}

$ =\frac{1}{2}d\sqrt{d^2+1}+\frac{1}{2}\int\frac{1}{\sqrt{d^2+1}}\mathrm d{d}

$ =\frac{1}{2}d\sqrt{d^2+1}+\frac{1}{2}\int\frac{1}{\sqrt{\sinh^2\theta+1}}\cdot(\cosh\theta)\mathrm d{\theta}

$ =\frac{1}{2}d\sqrt{d^2+1}+\frac{1}{2}\int\mathrm d{\theta}

$ =\frac{1}{2}d\sqrt{d^2+1}+\frac{1}{2}\theta+C

$ =\frac{1}{2}d\sqrt{d^2+1}+\frac{1}{2}\sinh^{-1}d+C

$ =\frac{1}{2}d\sqrt{d^2+1}+\frac{1}{2}\log(x+\sqrt{d^2+1})+C