1/х+1-1/х^2+4х+3=5/8

Решаем уравнение \[ \frac1{x+1}-\frac1{x^2+4x+3}=\frac58. \]

ОДЗ: \[ x\ne -1,\quad x\ne -3. \]

Так как \[ x^2+4x+3=(x+1)(x+3), \] то \[ \frac1{x+1}-\frac1{(x+1)(x+3)} =\frac{x+2}{(x+1)(x+3)}. \]

Получаем: \[ \frac{x+2}{(x+1)(x+3)}=\frac58. \]

\[ 8(x+2)=5(x+1)(x+3), \] \[ 8x+16=5x^2+20x+15, \] \[ 5x^2+12x-1=0. \]

\[ x=\frac{-12\pm\sqrt{164}}{10} =\frac{-6\pm\sqrt{41}}5. \]

Ответ: \[ \boxed{x_1=\frac{-6+\sqrt{41}}5,\qquad x_2=\frac{-6-\sqrt{41}}5}. \]