Visitor

2015-10-24 11:30:23

#1

a)?ln(2x+5)dx

b)?(1-2x)cosx/3dx

b)?(1-2x)cosx/3dx

Visitor

2015-10-24 11:45:33

#2

∫ln(2x+5)dx

∫(1-2x)cosx/3dx;

*This is what I wanted to write.*

∫(1-2x)cosx/3dx;

Visitor

2015-10-24 12:00:20

#3

Could be solved by integration by parts

f(x)=ln(2x+5)

f’(x)=2/(2x+5)

g(x)=x

g’(x)=1

F(x) = f(x) • g(x) - ∫g(x) • f’(x)dx

F(x) = x • ln(2x+5) - ∫x/(2x+5)dx

At the numerator we add and we cut 5 and we write

F(x) = x • ln(2x+5)-∫dx +5∫1/(2x+5)dx=

= x • ln(2x+5)-x+5ln(2x+5)+c

f(x)=ln(2x+5)

f’(x)=2/(2x+5)

g(x)=x

g’(x)=1

F(x) = f(x) • g(x) - ∫g(x) • f’(x)dx

F(x) = x • ln(2x+5) - ∫x/(2x+5)dx

At the numerator we add and we cut 5 and we write

F(x) = x • ln(2x+5)-∫dx +5∫1/(2x+5)dx=

= x • ln(2x+5)-x+5ln(2x+5)+c

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