Задача 33302 ...

leydi09

5 февраля 2019 г. в 22:20

5. ∫ (2 + 3x) / (x² – 5x – 3) dx

7. ∫ x arcctg 3x dx

sova

5 февраля 2019 г. в 22:34

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5.
x²–5x–3=(x²–2·(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/(t²–(37/4))=

=(3/2) ∫ d(t²–(37/4))/(t²–(37/4)) +(19/2) ∫ dt/(t²–(37/4))=

=(3/2)ln|t²–(37/4)| +(19/2)·(1/√37)ln|(2t–√37/(2t+√37|+C

=(3/2)ln|x²–5x–3|+(19/(2√37))ln|(2x–5–√37)/(2x–5+√37)|+C

7.
По частям
u=arcctgx
du=–dx/(1+x²)
dv=xdx
x=(x²)/2

=u·v– ∫ vdu= (x²·arcctgx)/2 + (1/2) ∫x²dx/(1+x²) =

=(x²·arcctgx)/2 + (1/2) ∫(x²+1–1)dx/(1+x²) =

=(x²·arcctgx)/2 + (1/2) ∫dx– (1/2) ∫ dx/(1+x²)=

=(x²·arcctgx)/2 + (1/2)x +(1/2) arcctgx + C