całka z arctgx

[renatkan]

\(\displaystyle{ \int}\) \(\displaystyle{ x^{2}arctg2xdx}\)

[sea_of_tears]

\(\displaystyle{ \int x^2arctg2x dx =
\begin{cases}
u=arctg2x & v^{'}=x^2 \\
u^{'}=\frac{1}{1+4x^2} & v=\frac{x^3}{3}
\end{cases}
\newline
=\frac{x^3}{3}arctg2x-\int \frac{x^3}{3}\cdot \frac{1}{1+4x^2}dx=
\frac{x^3}{3}arctg2x -\frac{1}{3}\int\frac{x^3}{1+4x^2}dx=\newline
=\frac{x^3}{3}arctg2x -\frac{1}{3}\cdot\frac{1}{4}\int\frac{4x^3}{1+4x^2}dx=
\frac{x^3}{3}arctg2x-\frac{1}{12}\int\frac{x(1+4x^2)-x}{1+4x^2}dx=\newline
\frac{x^3}{3}arctg2x-\frac{1}{12}(\int\frac{x(1+4x^2)}{1+4x^2}dx-\frac{x}{1+4x^2}dx)=\newline
\frac{x^3}{3}arctg2x-\frac{1}{12}(\int x dx - \frac{1}{8}\int\frac{8x}{1+4x^2}dx)=\newline
\frac{x^3}{3}arctg2x -\frac{1}{12}[\frac{x^2}{2}-\frac{1}{8}ln(1+4x^2)] +c}\)

[ Dodano: 23 Grudnia 2008, 23:11 ]
jakoś tak to powinno być, posprawdzaj sobie wszystko i wymnóż ten nawias jeszcze