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IvanOsipov
@IvanOsipov
July 2022
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Прошу, помогите с интегралами, пожалуйста..
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sedinalana
Verified answer
1
t=2+e^x ,dt=e^xdx
S[e^x/(2+e^x)dx=Sdt/t=lnt=ln(2+e^x)+C
2
u=lnx,du=dx/x
dv=dx/x,v=-1/2x²
Slnx/(x³)dx=-lnx/(2x²)+1/2Sdx/x³=-lnx/(2x²)-1/(4x²)+C
3
S(e^x+3^x)dx=e^x+3^x/ln3+C
4
u=lnx,du=dx/x
dv=xdx,v=x²/2
Sxlnxdx=1/2*x²*lnx-1/2Sxdx=1/2*x²*lnx-x²/4+C
5
S√(2x+5)dx=1/3*√(2x+5)³+C
6
S√(2+x)dx=2/3*√(2+x)³|4-(-1)=2/3*6√6-2/3*1=4√6-2/3
2 votes
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IvanOsipov
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IvanOsipov
Спасибо огромное!
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Answers & Comments
Verified answer
1t=2+e^x ,dt=e^xdx
S[e^x/(2+e^x)dx=Sdt/t=lnt=ln(2+e^x)+C
2
u=lnx,du=dx/x
dv=dx/x,v=-1/2x²
Slnx/(x³)dx=-lnx/(2x²)+1/2Sdx/x³=-lnx/(2x²)-1/(4x²)+C
3
S(e^x+3^x)dx=e^x+3^x/ln3+C
4
u=lnx,du=dx/x
dv=xdx,v=x²/2
Sxlnxdx=1/2*x²*lnx-1/2Sxdx=1/2*x²*lnx-x²/4+C
5
S√(2x+5)dx=1/3*√(2x+5)³+C
6
S√(2+x)dx=2/3*√(2+x)³|4-(-1)=2/3*6√6-2/3*1=4√6-2/3