Considering int(ln(x+1))/(2(x+1))dx I first solved it seeing it similar to the derivative of ln^2(x+1) so multiplying by 2/2 the solution is int(ln(x+1))/(2(x+1))dx=(ln^2(x+1))/(4)+const.. But then we can solve it using by parts' method and so this is the solution that I found: 1/2 int(ln(x+1))/((x+1))dx=1/2 ln(x+1)ln(x+1)-1/2 int(ln(x+1))/(x+1)dx

Tarnayfu

Tarnayfu

Open question

2022-08-16

Two different solutions of the same integral
Considering
ln ( x + 1 ) 2 ( x + 1 ) d x
I first solved it seeing it similar to the derivative of ln 2 ( x + 1 ) so multiplying by 2 2 the solution is
ln ( x + 1 ) 2 ( x + 1 ) d x = ln 2 ( x + 1 ) 4 + c o n s t .
. But then we can solve it using by parts' method and so this is the solution that I found:
1 2 ln ( x + 1 ) ( x + 1 ) d x = 1 2 ln ( x + 1 ) ln ( x + 1 ) 1 2 ln ( x + 1 ) x + 1 d x
Seeing it as an equation I brought the integral 1 2 ln ( x + 1 ) x + 1 d x to the left so that I obtain
ln ( x + 1 ) ( x + 1 ) d x = 1 2 ln ( x + 1 ) ln ( x + 1 ) + c o n s t .
so
ln ( x + 1 ) ( x + 1 ) d x = 1 2 ln 2 ( x + 1 ) + c o n s t .
. I know that the first solution is correct but I used to way of solution that seem to be correct. How is it possible? Where is the mistake? Thank you in advance for your help!

Answer & Explanation

pelvogp

pelvogp

Beginner2022-08-17Added 18 answers

Both answers are the same. Can you see that
l n ( x + 1 ) ( x + 1 ) d x = 1 2 ln ( x + 1 ) l n ( x + 1 ) + c o s t .
1 2 l n ( x + 1 ) ( x + 1 ) d x = 1 4 ln ( x + 1 ) l n ( x + 1 ) + c o s t .
Moselq8

Moselq8

Beginner2022-08-18Added 4 answers

Divide the last step by 2 to get the desired answer, as 2 c o n s t a n t = c o n s t a n t

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