How to evaluate int_0^1 ln(x+1)/(x^2+1)dx

Uroskopieulm

Uroskopieulm

Answered question

2022-11-05

How to evaluate 0 1 ln ( x + 1 ) x 2 + 1 d x

Answer & Explanation

motylowceyvy

motylowceyvy

Beginner2022-11-06Added 19 answers

Let x = tan ( t ). We then have d x = sec 2 ( t ) d t and
1 1 + x 2 = 1 1 + tan 2 ( t ) = 1 sec 2 ( t )
Hence, d x 1 + x 2 = d t
0 1 ln ( 1 + x ) 1 + x 2 d x = 0 π / 4 ln ( 1 + tan ( t ) ) d t = 0 π / 4 ln ( cos ( t ) + sin ( t ) ) d t 0 π / 4 ln ( cos ( t ) ) d t = 0 π / 4 ln ( 2 cos ( t π / 4 ) ) d t 0 π / 4 ln ( cos ( t ) ) d t = 0 π / 4 ln ( 2 ) d t + 0 π / 4 ln ( cos ( t π / 4 ) ) d t 0 π / 4 ln ( cos ( t ) ) d t = π ln ( 2 ) 8 + π / 4 0 ln ( cos ( t ) ) d t 0 π / 4 ln ( cos ( t ) ) d t = π ln ( 2 ) 8 ( cos ( t )  is even )
Hallie Stanton

Hallie Stanton

Beginner2022-11-07Added 5 answers

Substitute x = tan ( θ ) the integral then reduces to
0 π 4 ln ( 1 + tan ( θ ) ) d θ = 0 π 4 ln ( 2 cos ( π / 4 θ ) cos ( θ ) ) d θ = 0 π 4 ln ( 2 ) d θ = π ln ( 2 ) 8

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