Question

Integrala din arctg^6 din x/ 1+x^2

Answer 1

$\displaystyle \int \dfrac{\mathrm{artcg}^6x}{1+x^2}\, dx=\int \dfrac{1}{1+x^2}\cdot \mathrm{artcg}^6x\,\,dx=\\ \\ = \int (\mathrm{artcg}\,x)'\cdot \mathrm{artcg}^6x\,\, dx = \\ \\ =\dfrac{\mathrm{artcg}^{6+1}x}{6+1} +C =\\ \\ = \dfrac{\mathrm{artcg}^7x}{7}+C$