Задача 33302 ...

sova

2019-02-05 22:34:57

★

5.
x^2-5x-3=(x^2-2*(5/2)x+(25/4))-(25/4)-3=(x-(5/2))^2-(37/4)
Замена
x-(5/2)=t
x=t+(5/2)
dx=dt
2+3x=2+3*(t+(5/2)= 3t+(19/2)

= ∫ (3t+(19/2))dt/(t^2-(37/4))=

=(3/2) ∫ d(t^2-(37/4))/(t^2-(37/4)) +(19/2) ∫ dt/(t^2-(37/4))=

=(3/2)ln|t^2-(37/4)| +(19/2)*(1/sqrt(37))ln|(2t-sqrt(37)/(2t+sqrt(37)|+C

=(3/2)ln|x^2-5x-3|+(19/(2sqrt(37)))ln|(2x-5-sqrt(37))/(2x-5+sqrt(37))|+C

7.
По частям
u=arcctgx
du=-dx/(1+x^2)
dv=xdx
x=(x^2)/2

=u*v- ∫ vdu= (x^2*arcctgx)/2 + (1/2) ∫x^2dx/(1+x^2) =

=(x^2*arcctgx)/2 + (1/2) ∫(x^2+1-1)dx/(1+x^2) =

=(x^2*arcctgx)/2 + (1/2) ∫dx- (1/2) ∫ dx/(1+x^2)=

=(x^2*arcctgx)/2 + (1/2)x +(1/2) arcctgx + C