Найти определенный интеграл

Решите задачу:

$1)\; \int _0^1\frac{8arctgx-x}{1+x^2}dx=8\int _0^1arctgx\cdot \frac{dx}{1+x^2}-\frac{1}{2}\int _0^1\frac{2xdx}{1+x^2}=\\\\=[t=arctgx,\; u=1+x^2}]=8\cdot \int _0^{\frac{\pi}{4}}t\cdot dt-\frac{1}{2}\int _1^2\frac{du}{u}=8\frac{t^2}{2}|_0^{\frac{\pi}{4}}-\frac{1}{2}ln|u||_1^2\\\\=4\cdot \frac{\pi ^2}{16}-\frac{1}{2}(ln2-ln1)=\frac{\pi ^2}{4}-\frac{1}{2}ln2\\\\\\2)\; \int _1^{e}\frac{x^8+lnx^9+1}{x}dx=\int _1^{e}(x^7+\frac{9lnx}{x}+\frac{1}{x})dx=(\frac{x^8}{8}+9\frac{ln^2x}{2}+ln|x|)|_1^{e}=$

$=\frac{e^8}{8}+\frac{9}{2}+1-\frac{1}{8}$