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quasi
Posts:
10,576
Registered:
7/15/05


Re: Compute integral
Posted:
May 2, 2014 3:26 AM


cootercrew wrote: > >Compute the integral from 0 to 1 of ln(1+x)/(1+x^2)dx.
If the integrand was intended to be
ln((1+x)/(1+x^2))
then you need more parentheses, so as not to confuse it with
(ln(1+x))/(1+x^2)
I'll assume you meant the first of the above.
>Hint: Use the substitution y+1=2/(x+1)
I don't see how to use the hint, but I'd approach it this way:
int [0..1] (ln(1+x)/(1+x^2)) dx
= int [0..1] (ln(1+x)  ln(1+x^2)) dx
= (int [0..1] ln(1+x) dx)  (int [0..1] ln(1+x^2) dx)
Do each of the above integrals separately.
For the first integral, use integration by parts:
int [0..1] ln(1+x) dx
= u*v [0..1]  (int [0..1] v*du)
[where u = ln(1+x), v = x]
= x*ln(1+x) [0..1]  (int [0..1] x/(1+x) dx)
= ln(2)  (int [0..1] x/(1+x) dx)
Side work:
int [0..1] x/(1+x) dx
= int [0..1] (1  (1/(1+x))) dx
= (x  ln(1+x)) [0..1]
= 1  ln(2)
back to main ...
= ln(2)  (1  ln(2))
= 2*ln(2)  1
For the second integral, again use integration by parts:
int [0..1] (ln(1+x^2) dx
= u*v [0..1]  (int [0..1] v du)
[where u = ln(1+x^2), v = x]
= x*ln(1+x^2) [0..1]  (int [0..1] (2x^2)/(1+x^2) dx)
= ln(2)  (int [0..1] (2x^2)/(1+x^2) dx)
Side work:
int [0..1] (2x^2)/(1+x^2) dx
= int [0..1] (2  (2/(1+x^2)) dx = 2x [0..1]  2*arctan(x) [0..1]
= 2  2*(Pi/4)
= 2  Pi/2
back to main ...
= ln(2)  (2  Pi/2)
= ln(2)  2 + Pi/2
Combining the results yields
(2*ln(2)  1)  (ln(2)  2 + Pi/2)
= ln(2)  Pi/2 + 1
Remark:
If this is a practice problem for a Calculus II test, I wouldn't worry. Assuming a fair test, the above problem requires, in my opinion, too much brute force work for a single test problem.
quasi



