Question

integral de arc tg 3x xdx​

Answer 1

$\int {arctg(3x)\, x}\, dx$

Fórmulas:

• Derivada de arctg(ax):

$[arctg(ax)]'=\frac{a}{(ax)^2+1}$

• Integral por partes:

$\int{f(x)\, g'(x)\, dx=f(x)\, g(x)-\int{f'(x)\, g(x)\, dx$

Resolução:

$\int {arctg(3x)\, x}, dx\\\\=\int {arctg(3x)\, (\frac{x^2}{2})' }, dx\\\\={arctg(3x)\, \frac{x^2}{2}-\int{(arctg(3x))'\, \frac{x^2}{2}\, dx$

Pra facilitar, vou focar apenas na segunda integral e depois juntar as respostas;

$\frac{1}{2}\int{(arctg(3x))' x^2\, dx$

$=\frac{1}{2}\int{\frac{3}{(3x)^2+1} x^2\, dx$

$=\frac{3}{2}\int{\frac{x^2}{9x^2+1}\, dx$

$=\frac{1}{6}\int{\frac{9x^2}{9x^2+1}\, dx$

$=\frac{1}{6}\int{1-\frac{1}{9x^2+1}\, dx$

$=\frac{1}{6}\int{1}\, dx-\frac{1}{6}\int\ {\frac{1}{9x^2+1}\, dx$

$=\frac{1}{6}x-\frac{1}{18}\int\ {\frac{3}{(3x)^2+1}\, dx$

$=\frac{1}{6}x-\frac{1}{18}arctg(3x)+C$

Com isso, a resposta é:

$\int {arctg(3x)\, x}, dx\\={arctg(3x)\, \frac{x^2}{2}-(\frac{1}{6}x-\frac{1}{18}arctg(3x))+C$

$=\frac{1}{18} arctg(3x)(9x^2+1)-\frac{1}{6}x+C$

Se estiver com alguma dúvida, pode me chamar nos comentários. Bons estudos ^~