Question

Интеграл.$intlimits , arctgsqrt{6x-1} dx$

Answer 1

$intarctan(sqrt{6x-1})dx=begin{vmatrix}t=6x-1\dt=6dxend{vmatrix}=dfrac{1}{6}intarctan(sqrt{t})du=begin{vmatrix}y=sqrt{t}\dy=dfrac{1}{2sqrt{t}}dtend{vmatrix}=dfrac{1}{3}int yarctan(y)dy=begin{vmatrix}u=arctan(y)&dv=ydy\du=dfrac{1}{y^2+1}dy&v=dfrac{y^2}{2}end{vmatrix}=dfrac{y^2arctan(y)}{6}-dfrac{1}{6}intdfrac{y^2}{y^2+1}dy=dfrac{y^2arctan(y)}{6}-dfrac{1}{6}intleft(1-dfrac{1}{y^2+1}right)dy=dfrac{y^2arctan(y)}{6}+dfrac{arctan(y)}{6}-dfrac{y}{6}+C=$

$dfrac{(6x-1)arctan(sqrt{6x-1})}{6}+dfrac{arctan(sqrt{6x-1})}{6}-dfrac{sqrt{6x-1}}{6}+C=xarctan(sqrt{6x-1}+dfrac{sqrt{6x-1}}{6}+C$