Question

Resolva as integrais a seguir usando o método de integração por partes.

Answer 1

1)

a)

∫ x * e^(x)  dx

u=x  ==> du =dx

dv=e^(x)  dx   ==> ∫dv=∫e^(x)  dx  ==>v=e^(x)

∫ x * e^(x)  dx =x* e^(x)- ∫e^(x) dx

∫ x * e^(x)  dx =x* e^(x)- e^(x)  + c

b)

∫ x * ln(x)  dx

u= ln(x)  ==>du= dx/x

dv =x dx ==>∫ dv =  ∫ x  dv  ==> v= x²/2

∫ x * ln(x)  dx= (x²/2) * ln(x)- ∫ x²/2   dx/x

∫ x * ln(x)  dx= (x²/2) * ln(x)- (1/2)*∫  x  dx

∫ x * ln(x)  dx= (x²/2) * ln(x)- (1/2)*x²/2 + c

∫ x * ln(x)  dx= (x²/2) * ln(x)- (1/4)*x²+ c

c)

∫ x² * ln(x)  dx

u=ln(x)  ==> du=dx/x

dv=x²   dx    ==>∫ dv=∫x²   dx ==>v= x³/3

∫ x² * ln(x)  dx = ln(x) * x³/3 - ∫ x³/3  * dx/x

∫ x² * ln(x)  dx = ln(x) * x³/3 -(1/3)* ∫ x²  * dx

∫ x² * ln(x)  dx = ln(x) * x³/3 -(1/3)* x³/3  + c

∫ x² * ln(x)  dx = ln(x) * x³/3 -(1/9)* x³  + c

d)

∫  x * sec²(x) dx

u=x ==> du=dx

du =  sec²(x) dx ==>∫ du = ∫ sec²(x) dx ==>u =tan(x)

∫  x * sec²(x) dx = x * tan(x) - ∫ tan(x) dx

∫  x * sec²(x) dx = x * tan(x) - (-ln(cos(x))) +c

∫  x * sec²(x) dx = x * tan(x) +ln(cos(x)) +c

e)

∫ √x * ln (x) dx

u=ln (x)  ==> du=dx/x

dv =  √x  dx  ==>∫  dv =  ∫ √x  dx ==> v= (2/3) * x^(3/2)

∫ √x * ln (x) dx =  (2/3) * x^(3/2) * ln(x) - ∫(2/3) * x^(3/2) dx/x

∫ √x * ln (x) dx =  (2/3) * x^(3/2) * ln(x) -(2/3)* ∫√x dx

∫ √x * ln (x) dx =  (2/3) * x^(3/2) * ln(x) -(2/3)* x^(1/2+1)/(1/2+1) +c

∫ √x * ln (x) dx =  (2/3) * x^(3/2) * ln(x) -(2/3)* √x³/(3/2) +c

∫ √x * ln (x) dx =  (2/3) * x^(3/2) * ln(x) - (4/9)* √x³+c

f)

∫(x+1)* cos(2x) dx

=∫x* cos(2x) dx + ∫ cos(2x) dx

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∫x* cos(2x) dx

u =x ==> du = dx

dv =  cos(2x) dx   ==> ∫ dv=∫cos(2x) dx ==> v= (1/2) * sen(2x)

∫x* cos(2x) dx= (x/2) * sen(2x) - ∫ (1/2) * sen(2x)  dx

∫x* cos(2x) dx= (x/2) * sen(2x) - (1/2)*(-1/2) * cos(2x)) + c

∫x* cos(2x) dx= (x/2) * sen(2x) +(1/4) * cos(2x)) + c

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∫ cos(2x) dx =  (1/2) * sen(2x)  + c

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∫(x+1)* cos(2x) dx =  (x/2) * sen(2x) +(1/4) * cos(2x))   +   (1/2) * sen(2x)  + c