Question

Помогите вычислить определенный интеграл

Answer 1

$\displaystyle \int\limits^{0,25}_0 arctg4xdx=xarctg4x|^{0,25}_0-4\int\limits^{0,25}_0\frac{xdx}{1+16x^2}=xarctg4x|^{0,25}_0-\\-\frac{1}{8}\int\frac{d(1+16x^2)}{1+16x^2}=xarctg4x|^{0,25}_0-\frac{1}{8}ln|1+16x^2||^{0,25}_0=\\=\frac{1}{4}arctg1-\frac{1}{8}ln2\approx0,110\\u=arctg4x;du=\frac{4dx}{1+16x^2}\\dv=dx;v=x$

$\displaystyle\int\limits^{0,25}_0\frac{4xdx}{1+16x^2}=\frac{1}{8}\int\limits^2_1\frac{dt}{t}=\frac{1}{8}ln|t||^2_1=\frac{1}{8}ln2\\1+16x^2=t\\dt=32xdx;dx=\frac{dt}{32x}\\t_1=1+16*\frac{1}{16}=2\\t_2=1$