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integrala definita

adrian
Vizitator
2016-02-12 18:06:51

Buna ziua


Am de calculai urmatoarea integrala definita:



Problema ar fi simpla daca functia ar fi impara  dar cred ca nu este.


Atunci?O sugestie multumescSmile

Adrian
Vizitator
2016-02-23 22:45:29

Buna seara


cerinta nu mai este de actualitate am rezolvat intre timp integrala.

Coanda Ilie
Vizitator
2020-08-12 17:27:23
Problem:

x2ex+1dx

Substitute u=ex dudx=ex (steps) dx=exdu, use:
x2=ln2(u)

=ln2(u)u(u+1)du

Perform partial fraction decomposition:

=(ln2(u)uln2(u)u+1)du

Apply linearity:

=ln2(u)uduln2(u)u+1du



Now solving:

ln2(u)udu

Substitute v=ln(u) dvdu=1u (steps) du=udv:

=v2dv

Apply power rule:

vndv=vn+1n+1 with n=2:

=v33

Undo substitution v=ln(u):

=ln3(u)3



Now solving:

ln2(u)u+1du

Substitute v=u+1 dvdu=1 (steps) du=dv:

=ln2(v1)vdv

Integrate by parts in order to solve using polylogarithms:

=ln2(v1)ln(v)2ln(v1)ln(v)v1dv



Now solving:

ln(v1)ln(v)v1dv

Integrate by parts in order to solve using polylogarithms:

=Li2(1v)ln(v1)+Li2(1v)v1dv



Now solving:

Li2(1v)v1dv

Substitute w=1v dwdv=1 (steps) dv=dw:

=Li2(w)wdw

Simplify:

=Li2(w)wdw

This is a standard integral:

=Li3(w)

Undo substitution w=1v:

=Li3(1v)



Plug in solved integrals:

Li2(1v)ln(v1)+Li2(1v)v1dv

=Li3(1v)Li2(1v)ln(v1)



Plug in solved integrals:

ln2(v1)ln(v)2ln(v1)ln(v)v1dv

=ln2(v1)ln(v)+2Li2(1v)ln(v1)2Li3(1v)

Undo substitution v=u+1:

=ln2(u)ln(u+1)+2Li2(u)ln(u)2Li3(u)



Plug in solved integrals:

ln2(u)uduln2(u)u+1du

=ln2(u)ln(u+1)+ln3(u)32Li2(u)ln(u)+2Li3(u)

Undo substitution u=ex, use:
ln(ex)=x

=x2ln(ex+1)+2Li3(ex)2xLi2(ex)+x33



The problem is solved:

x2ex+1dx

=x2ln(ex+1)+2Li3(ex)2xLi2(ex)+x33+C

Rewrite/simplify:

=x(x(3ln(ex+1)x)+6Li2(ex))6Li3(ex)3+C
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