integration

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\hspace{-16}$If Integral is $\bf{\int \sqrt{x+\sqrt{x^2+2}}dx}$\\\\\\ Let $\bf{x+\sqrt{x^2+2}=t^2\Leftrightarrow \left(\frac{x+\sqrt{x^2+2}}{\sqrt{x^2+2}}\right)dx=2tdt}$\\\\\\ Now $\bf{\left(\sqrt{x^2+2}+x\right).\left(\sqrt{x^2+2}-x\right)=2}$\\\\\\ So $\bf{\sqrt{x^2+2}-x=\frac{2}{t^2}}$\\\\\\ So $\bf{2\sqrt{x^2+2}=t^2+\frac{2}{t^2}\Leftrightarrow \left(\sqrt{x^2+2}\right)=\frac{t^4+2}{2t^2}}$\\\\\\ So $\bf{\frac{2t^2.t^2}{t^4+2}dx=2tdt\Leftrightarrow dx=\frac{t^4+2}{t^3}dt}$\\\\\\ So $\bf{\int \frac{(t^4+2)}{t^3}.tdt=\int \frac{t^4+2}{t^2}dt}$\\\\\\ So $\bf{\int t^2dt+\int \frac{2}{t^2}dt}$\\\\\\ $\bf{=\frac{1}{3}.t^3-2.\frac{1}{t}+\mathbb{C}}$\\\\\\ So $\bf{\sqrt{x+\sqrt{x^2+2}}dx=\frac{1}{3}.\left(x+\sqrt{x^2+2}\right)^3-2.\frac{1}{\left(x+\sqrt{x^2+2}\right)}+\mathbb{C}}$