Задача 57390 Прошу помогите хоть с каким нибудь,...
Условие
Решение
4dx=dt => dx=[m]\frac{dt}{4}[/m], получаем
[m]\int{\frac{dt}{4t^{\frac{1}{3}}}}=\frac{1}{4}\int{t^{-\frac{1}{3}}dt}=\frac{1}{4}*\frac{3t^\frac{2}{3}}{2}=\frac{3t^\frac{2}{3}}{8}+c=\frac{3(4x+1)^\frac{2}{3}}{8}+c[/m]
[red]2)[/red][m]\int{\frac{(3x+1)^\frac{1}{4}-(3x+1)^\frac{2}{5}}{(3x+1)^\frac{2}{5}}}dx=\int{\frac{(3x+1)^\frac{1}{4}(1-(3x+1)^\frac{3}{20})}{(3x+1)^\frac{2}{5}}}dx=\int{\frac{1-(3x+1)^\frac{3}{20})}{(3x+1)^\frac{3}{20}}}dx=\int{(\frac{1}{(3x+1)^\frac{3}{20}}-1)dx}=\int{\frac{dx}{(3x+1)^{\frac{3}{20}}}}-\int{dx}[/m]
[green]решаем первый интеграл: [/green]
замена: 3x+1=t => 3dx=dt => dx=[m]\frac{dt}{3}[/m]
[m]\int{\frac{dx}{(3x+1)^{\frac{3}{20}}}} = \int{\frac{dt}{3t^\frac{3}{20}}}=\frac{1}{3}\int{t^-\frac{3}{20}dt}=\frac{1}{3}*\frac{20t^\frac{17}{20}}{17}=\frac{20t^\frac{17}{20}}{51}+c=\frac{20(3x+1)^\frac{17}{20}}{51}+c[/m]
[green]решаем второй интеграл:[/green]
[m]\int{dx}=x+c[/m]
[green]собираем всё вместе:[/green]
[m]\int{\frac{dx}{(3x+1)^{\frac{3}{20}}}}-\int{dx}=\frac{20(3x+1)^\frac{17}{20}}{51}-x+c[/m]
[red]3)[/red]x^(2)-4x+7=(x^(2)-2*2*x+4)+3 =(x-2)^(2)+3
[m]\int{\frac{dx}{(x-2)^2+3}}=\int{\frac{dx}{(x-2)^2+(\sqrt{3})^2}} =\frac{1}{\sqrt{3}}arctan(\frac{x-2}{\sqrt{3}})+c [/m]
[red]4)[/red][m]\int{\frac{dx}{\sqrt{3-x-x^2}}}=\int{\frac{dx}{\sqrt{-(x^2+x-3)}}}=\int{\frac{dx}{\sqrt{-(x^2+x+\frac{1}{4}-\frac{13}{4})}}}=\int{\frac{dx}{\sqrt{-((x+\frac{1}{2})^2-\frac{13}{4})}}}=\int{\frac{dx}{\sqrt{\frac{13}{4}-(x+\frac{1}{2})^2}}} = \int{\frac{dx}{\sqrt{(\sqrt{\frac{13}{4}})^2-(x+\frac{1}{2})^2}}} = arcsin(\frac{x+\frac{1}{2}}{\sqrt{\frac{13}{4}}})+c[/m]
[red]5)[/red][m]\int{\frac{dx}{\sqrt{1-(x-1)^2}}}=arcsin(\frac{x-1}{1})+c=arcsin(x-1)+c[/m]
[red]6)[/red][m]\int{\frac{dx}{8+4x^2}}=\int{\frac{dx}{4x^2+8}}=\int{\frac{dx}{4(x^2+2)}}=\frac{1}{4}\int{\frac{dx}{x^2+2}}=\frac{1}{4}\int{\frac{dx}{x^2+(\sqrt{2})^2}}=\frac{1}{4}*\frac{1}{\sqrt{2}}arctan(\frac{x}{\sqrt{2}})=\frac{1}{4\sqrt{2}}arctan(\frac{x}{\sqrt2})+c[/m]
[red]7)[/red][m]\int{(1-ctg(4x))dx}=\int{dx}-\int{ctg(4x)dx}=\int{dx}-\int{\frac{cos(4x)}{sin(4x)}dx} [/m]
[green]Решаем второй интеграл: [/green]
сделаем замену переменной: пусть sin(4x)=t => 4cos(4x)dx=dt => dx=[m]\frac{dt}{4cos(4x)}[/m]
[m]\int{\frac{cos(4x)}{sin(4x)}dx}=\int{\frac{cos(4x)}{t}*\frac{dt}{4cos(4x)}}=\frac{1}{4}\int{\frac{dt}{t}}=\frac{ln|t|}{4}+c=\frac{ln|sin(4x)|}{4}+c[/m]
[green]Решаем первый интеграл:[/green]
[m]\int{dx}=x+c[/m]
[green]собираем решение:[/green]
[m]\int{dx}-\int{\frac{cos(4x)}{sin(4x)}dx} =x-\frac{ln|sin(4x)|}{4}+c[/m]
[red]8)[/red]сделаем замену: 2x+3=t => 2dx=dt => dx=[m]\frac{dt}{2}[/m]
[m]\int{\frac{dt}{2t}}=\frac{1}{2}\int{\frac{dt}{t}}=\frac{ln|t|}{2}+c=\frac{ln|2x+3|}{2}+c[/m]