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Topic: Compute integral
Replies: 7   Last Post: May 2, 2014 1:36 PM

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 quasi Posts: 12,012 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

Date Subject Author
5/2/14 Jeremy Johnson
5/2/14 Math Lover
5/2/14 Roland Franzius
5/2/14 quasi
5/2/14 Jeremy Johnson
5/2/14 quasi
5/2/14 Peter Percival
5/2/14 Virgil