Question

$\int\frac{\sqrt{arctg^{3}*5x }-3 }{1+25x^{2} } \, dx$Вычислить интегал.

Помогите пожалуйста!!!

Answer 1

Ответ:

$\int \dfrac{\sqrt{arctg^35x}-3}{1+25x^2}\, dx=\int \dfrac{\sqrt{arctg^35x}}{1+25x^2}\, dx-\int \dfrac{3\, dx}{1+25x^2}=\\\\\\=\Big[\ t=arctg5x\ ,\ dt=\dfrac{5\, dx}{1+25x^2}\ \Big]=\dfrac{1}{5}\, \int t^{3/2}\, dt-\dfrac{3}{5}\, \int \dfrac{5dx}{1+(5x)^2}=\\\\\\=\dfrac{1}{5}\cdot \dfrac{t^{5/2}}{5/2}-\dfrac{3}{5}\cdot arctg5x+C=\dfrac{2}{25}\cdot \sqrt{arctg^55x}-\dfrac{3}{5}\cdot arctg5x+C$