freak integral

I=\int {\frac{1}{({x+\sqrt{x^2-1}})^2}}dx \\=\int {({x-\sqrt{x^2-1}})^2}dx \\=\int {2x^2-1-2x\sqrt{x^2-1}}dx \\=\frac {2}{3}x^3-x-2\int {x\sqrt{x^2-1}}dx \\now, \\z^2=x^2-1 \\zdz=xdx \\I=\frac {2}{3}x^3-x-2\int z^2dz \\=\frac {2}{3}x^3-x-\frac{4}{3}z^3+C \\=\frac {2}{3}x^3-x-\frac{4}{3}{(\sqrt{x^2-1})}^3+C

thnks sir but i had solved the question earlier.just posted for practice.
anyways thnks