Найдите неопределенные интегралы. результат проверить дифференцированием. 112 задание

Question 1

Question 2

$\int{\sqrt{(2-x^2)^3}x}\, dx=\frac12\int{\sqrt{(2-x^2)^3}}\, dx^2=\frac15(2-x^2)^{\frac52}$

Проверка:

$\frac{d(\frac25(2-x^2)^{\frac52})}{dx^2}}=\frac12*\frac25*\frac52*(2-x^2)^{\frac32}*2=$

$=(2-x^2)^{\frac32}$

Question 3

$112_c \ \int arctg(\sqrt{x}) \ dx = [\sqrt{x} = t, \ \frac{1}{2\sqrt{x}}dx = dt ] = \int 2t \ arctg(t) \ dt =\\\\t^2arctg(t) - \int t^2\frac{1}{1+t^2} \ dt = t^2arctg(t) - \int \frac{1 + t^2 - 1}{1 + t^2} \ dt =\\\\ t^2arctg(t) - \int 1 - \frac{1}{1 + t^2} \ dt = t^2arctg(t) - t + arctg(t) + C =\\\\ xarctg(\sqrt{x}) - \sqrt{x} + arctg(\sqrt{x}) + C = (x+1)arctg({\sqrt{x}}) - \sqrt{x} + C$

Проверка:

$((x+1)arctg({\sqrt{x}}) - \sqrt{x} + C)' = ((x+1)arctg({\sqrt{x}}))' - \frac{1}{2\sqrt{x}} =\\\\ (x+1)'*arctg({\sqrt{x}}) + (x+1)*(arctg({\sqrt{x}}))' - \frac{1}{2\sqrt{x}} =\\\\ arctg({\sqrt{x}}) + (x+1)*(\sqrt{x})'\frac{1}{1 + (\sqrt{x})^2} - \frac{1}{2\sqrt{x}} =$

$arctg({\sqrt{x}}) + (x+1)*\frac{1}{2\sqrt{x}}\frac{1}{1 + (\sqrt{x})^2}- \frac{1}{2\sqrt{x}}=\\\\ arctg({\sqrt{x}}) + \frac{x + 1}{2\sqrt{x}(1 + x)} - \frac{1}{2\sqrt{x}} =\\\\ arctg({\sqrt{x}}) + \frac{1}{2\sqrt{x}} - \frac{1}{2\sqrt{x}} = \boxed{ arctg({\sqrt{x}})}$

$112_a \ \int \sqrt{(2 - x^2)^3} x dx = [2 - x^2 = t, -2xdx = dt] = -\int \frac{\sqrt{t^3}}{2} \ dt =\\\\ -\int \frac{t^{\frac{3}{2}}}{2} \ dt = -\frac{1}{5}t^{\frac{5}{2}} + C = -\frac{1}{5}(2 - x^2)^{\frac{5}{2}} + C =-\frac{1}{5}\sqrt{(2 - x^2)^5} + C$

Проверка:

$(-\frac{1}{5}\sqrt{(2 - x^2)^5} + C)' = -\frac{(2 -x^2)'}{5}\frac{5}{2}\sqrt{(2-x^2)^3} = \frac{2x}{5}\frac{5}{2}\sqrt{(2-x^2)^3} =\\\\ \boxed{ x\sqrt{(2-x^2)^3} }$

$112_b \ \int \frac{6x - 7}{3x^2 - 7x + 11} \ dx = [3x^2 - 7x + 11 = t,(6x - 7)dx = dt] =\\\\ \int \frac{1}{t} \ dt = ln(t) + C = ln(3x^2 - 7x + 11) + C$

Проверка:

$(ln(3x^2 - 7x + 11) + C)' = (3x^2 - 7x + 11)'*\frac{1}{3x^2 - 7x + 11} =\\\\ \boxed{ \frac{6x - 7}{3x^2 - 7x + 11} }$