Задача 70101 ...
Условие
Решение
[m]u=arctg4x[/m] ⇒
[m]du=(arctg4x)`dx=\frac{1}{1+(4x)^2}\cdot (4x)`dx=\frac{4}{1+16x^2}dx[/m]
[m]dv=dx[/m] ⇒
[m]v= ∫dv= ∫ dx=x[/m]
[m] ∫ \underbrace{arctg4x}_{u} \underbrace{dx}_{dv}=\underbrace{ arctg4x}_{u}\cdot \underbrace{x}_{v}- ∫ \underbrace{x}_{v}\cdot \underbrace{\frac{4}{1+16x^2}dx}_{du}=x\cdot arctg4x - ∫ \frac{4x}{1+16x^2}dx=[/m]
так как
[m]d(1+16x^2)=(1+16x^2)`dx=32xdx[/m] ⇒
[m]4xdx=\frac{1}{8}d(1+16x^2)[/m]
и последний интеграл табличный : [r] [m] ∫ \frac{du}{u}=ln|u|[/m][/r]
Итак
[m]∫ \underbrace{arctg4x}_{u} \underbrace{dx}_{dv}=x\cdot arctg4x - \frac{1}{8}∫ \frac{(1+16x^2)}{1+16x^2}dx=x\cdot arctg4x - \frac{1}{8}ln(1+16x^2) + C[/m]