Question

найдите неопределенные интегралы. результат проверить дифференцированием. 112 задание

Answer 1

$int{sqrt{(2-x^2)^3}x}, dx=frac12int{sqrt{(2-x^2)^3}}, dx^2=frac15(2-x^2)^{frac52}$

 

Проверка:

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

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

Answer 2

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

 

 

Проверка:

 

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

 

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

 

 

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

 

 

Проверка:

 

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

 

 

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

 

 

Проверка:

 

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