oblicz calke

[gufox]

\(\displaystyle{ \int x ^{2}arcctgx}\)

[mol_ksiazkowy]

przez czesci tj =\(\displaystyle{ \frac{x^3}{3} arctg(x) - t \frac{x^3}{3(1+x^2)} dx}\)
te ostatnia podst
\(\displaystyle{ t=1+x^2}\)

[gufox]

dzieki, prosze o skasowanie postu, bo zdublowalem

[agulka1987]

\(\displaystyle{ \int x^2 arcctgx dx= \begin{bmatrix}f'=x^2& g=arcctgx\\f= \frac{1}{3}x^3& g' = \frac{-1}{1+x^2}\end{bmatrix} = \frac{1}{3}x^3 arcctgx + \frac{1}{3} t \frac{x^3}{1+x^2}dx = \frac{1}{3}x^3 arcctgx + \frac{1}{3} t x- \frac{x}{1+x^2}dx = \frac{1}{3}x^3 arcctgx + \frac{1}{3} t xdx - \frac{1}{3} t \frac{x}{1+x^2}dx = \frac{1}{3}x^3 arcctgx + \frac{1}{6} x^2 - \frac{1}{3} t \frac{x}{1+x^2}dx = \begin{bmatrix} t=1+x^2\\dt=2xdx\\ \frac{1}{2}st = xdx \end{bmatrix} = \frac{1}{3}x^3 arcctgx + \frac{1}{6} x^2 - \frac{1}{6} t \frac{1}{t}dt = \frac{1}{3}x^3 arcctgx + \frac{1}{6} x^2 - \frac{1}{6} ln|t| + C = \frac{1}{3}x^3 arcctgx + \frac{1}{6} x^2 - \frac{1}{6} ln|1+x^2| + C}\)

[gufox]

podobne bylo z arctgx, dzieki