Интеграл xdx/sqrt(2+3x-2x^2) Интеграл ((x-4)dx)/(5x^2-x+7)

Решите задачу:

$1)\; \; \int \frac{dx}{\sqrt{2+3x-2x^2}}=[\, 2+3x-2x^2=-2(x^2-\frac{3}{2}x-1)=\\\\=-2((x-\frac{3}{4})^2-\frac{9}{16}-1)=-2((x-\frac{3}{4})^2-\frac{25}{16})=2(\frac{25}{16}-(x-\frac{3}{4})^2\, ]=\\\\=\frac{1}{2}\int \frac{dx}{\sqrt{\frac{25}{16}-(x-\frac{3}{4})^2}}=[\, t=x-\frac{3}{4},\; dx=dt\, ]=\frac{1}{2}\int \frac{dt}{\sqrt{\frac{25}{16}-t^2}}=\\\\=\frac{1}{2}\cdot arcsin\frac{t}{5/4}+C=\frac{1}{2}\cdot arcsin\frac{4(x-\frac{3}{4})}{5}+C=\frac{1}{2}\cdot arcsin\frac{4x-12}{5}+C$

$2)\; \; \int \frac{(x-4)dx}{5x^2-x+7}=[\, 5(x^2-\frac{1}{5}x+\frac{7}{5})=5((x-\frac{1}{10})^2-\frac{1}{100}+\frac{7}{5})=\\\\=5((x-\frac{1}{10})^2-\frac{139}{100})\, ]=\frac{1}{5}\int \frac{(x-4)dx}{(x-0,1)^2-1,39}=[\, t=x-0,1\; ;\; dt=dx\, ]=\\\\=\frac{1}{5}\int \frac{(t-3,9)dt}{t^2-1,39}=\frac{1}{10}\int \frac{2t\, dt}{t^2-1,39}-\frac{3,9}{5}\int \frac{dt}{t^2-1,39}=\\\\=0,1\cdot ln|t^2-1,39|-\frac{39}{50}\cdot \frac{10}{2\sqrt{139}}\cdot ln\Big |\frac{t-\sqrt{1,39}}{t+\sqrt{1,39}}\Big |+C=\\\\=0,1\cdot ln|x^2-0,2x+1,4|-\frac{39}{2\sqrt{139}}\cdot ln\Big |\frac{10x-1-\sqrt{139}}{10x-1+\sqrt{139}}\Big |+C$