$x''-2x'+x=0;\:x\left(0\right)=0;\:x'\left(0\right)=1$

$25\:+\frac{25}{16}x^{2\:}\ge0$

$\int\frac{\left(4x+3\right)}{4x^4+8x^3+3x^2}dx$

$2\cdot\left(a+1\right)\cdot\left(2a-3\right)$

$\int e^{-3}dx$

$\frac{\left(6a+4a^2\right)}{\left(3a^2+1\right)}=\left(6a+a^2\right)$

$5,776\:-\:-8$