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EXERCÍCIOS M1

Anatolie Sochirca ACM DEETC ISEL

Cálculo de integrais indefinidos. Integração por partes.
Seja u (x) e v(x) funções diferenciáveis em I Se a função u ′( x) ⋅ v( x) tem primitiva em I então e a função u ( x) ⋅ v ′( x) tem primitiva em I e tem-se

∫ u( x) ⋅ v ′( x) ⋅ d x = u ( x) ⋅ v( x) − ∫ v( x) ⋅ u ′( x) ⋅ d x ou ∫ u ( x) ⋅ d v ( x ) = u ( x ) ⋅ v ( x ) − ∫ v ( x ) ⋅ d u ( x ) ; ∫ u ⋅ d v = u ⋅ v − ∫ v ⋅ d u .
►1)

∫ln x ⋅ d x .

Consideramos u = l n x e d v = d x .

1 ′ Então u = l n x ⇒ d u = d (l n x ) = (l n x ) ⋅ d x = ⋅ d x x e dv = d x ⇒ v = x . Portanto temos:

∫ l n x ⋅ d x= x ⋅ l n x − ∫ x ⋅ x ⋅ d x= x ⋅ l n x − ∫ d x= x ⋅ l n x − x + C .
►2)

1

■

∫x

2

⋅ ln x ⋅ d x .

Consideramos u = l n x e d v = x 2 ⋅ d x . 1 ′ Então u = l n x ⇒ d u = d (l n x ) = (l n x ) ⋅ d x = ⋅ d x x e x3 2 2 dv = x ⋅d x ⇒ v = ∫ x ⋅ d x = . 3 Portanto temos:
2 ∫ x ⋅ln x ⋅ d x =

x3 x3 1 x3 x2 ⋅ ln x − ∫ ⋅ ⋅ d x = ⋅ ln x − ∫ ⋅ d x = 3 3 x 3 3

=

1 x3 x3 x3 ⋅ln x − ⋅ ∫ x2 ⋅ d x = ⋅ ln x − +C. 3 3 3 9

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1

EXERCÍCIOS M1

Anatolie Sochirca ACM DEETC ISEL

►3)

∫ x ⋅ arctg x ⋅ d x .
1 ⋅dx 1+ x2

Consideramos u = arctg x e d v = x ⋅ d x .
′ Então u = arctg x ⇒ d u = d (arctg x ) = (arctg x ) ⋅ d x =

e x2 dv = x ⋅ d x ⇒ v = ∫ x⋅ d x = . 2

Portanto temos:

∫ x ⋅ arctg x ⋅ d x =
=

x2 x2 1 x2 1 x2 ⋅ arctg x − ∫ ⋅ ⋅dx = ⋅ arctg x − ∫ ⋅ ⋅dx = 2 2 1+ x2 2 2 1+ x2

x2 1 1+ x2 −1 x2 1 1+ x2 1 ⋅ arctg x − ⋅ ∫ ⋅dx = ⋅ arctg x − ⋅ ∫  2 1+ x2 − 1+ x2 2 2 2 2  1+ x

 ⋅dx =  

=

1  1  1 1 1 x2 x2 ⋅ arctg x − ⋅ ∫ 1 − ⋅ arctg x − ⋅ ∫ d x + ⋅ ∫ ⋅dx = ⋅dx = 2 2 2  1+ x  2 2 2 1+ x2 x2 1 1 x2 1 1 ⋅ arctg x − ⋅ x + ⋅ arctg x + C = ⋅ arctg x + ⋅ arctg x − ⋅ x + C . 2 2 2 2 2 2

=

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►4)

∫ x ⋅ arcsenx ⋅ d x .
1 1− x2 ⋅dx

Consideramos u = arcsen x e d v = x ⋅ d x .

′ Então u = arcsen x ⇒ d u = d (arcsen x ) = (arcsen x ) ⋅ d x = e dv = x ⋅ d x ⇒ v = ∫ x⋅ d x = Portanto temos: