Zadanie nr 9213280

Oblicz całkę ∫ ∘ 1−x- dx 1+x-⋅-x .

Rozwiązanie

Jeżeli podstawimy 2 1−x- t = 1+x to mamy

2 t(1 + x ) = 1− x x(t2 + 1) = 1− t2 2 x = 1-−-t- t2 + 1 −2t(t2 + 1) − (1− t2)⋅2t − 4t dx = ----------2-----2--------dt = --2----2-dt. (t + 1) (t + 1)

Mamy więc

∫ ∘ ------ ∫ 2 ∫ 2 1-−-x-⋅ dx-= t⋅ t-+-1-⋅---−-4t--dt = ------4t--------dt = 1 + x x 1− t2 (t2 + 1)2 (t2 + 1)(t2 − 1) ∫ ( 1 1 ) ∫ 2 = 2 ------+ ------ dt = 2 arctg t+ --------------dt = t2 + 1 ( t2 − 1 ) (t − 1)(t+ 1) ∫ 1 1 = 2 arctg t+ -----− ----- dt = 2 arctgt + ln|t− 1|− ln |t + 1|+ C = | t− 1| t+ 1 ∘ ------ | √ ------ √ -----| |t− 1| 1− x || 1− x − 1+ x|| = 2 arctg t+ ln ||-----||+ C = 2 arctg ------+ ln| √---------√------|+ C = t+ 1 1+ x | 1− x + 1+ x| ∘ ------ || √ ------ √ ------2 || = 2 arctg 1−-x-+ ln ||(--1-−-x-−---1-+-x)--||+ C 1+ x | − 2x | ∘ ------ √ ------- = 2 arctg 1−-x-+ ln 1-−---1-−-x-2+ C. 1+ x |x|

Odpowiedź: ∘ ---- √----2 2 arctg 1−1+xx--+ ln 1−-|1x−|x + C