∫dx/√x^5 = ∫x^(-5/2) dx = -(2/3)*x^(-3/2) = -2/(3*x(3/2)) + C
∫dx/(1+9x)dx Сделаем замену u = 1+9x; du = 9dx; dx = (1/9) *du
∫dx/(1+9x)dx = ∫(1/9)* du/u = (1/9) * ln(u) = (1/9) * ln(1+9x) + C
∫e^(5x-7)dx Сделаем замену u = 5x-7; du = 5dx; dx = (1/5)du
∫e^(5x-7)dx = ∫(1/5)*e^u du = (1/5) * e^u = (1/5) e^(5x-7) + C