Question

Помогите решить интеграл поэтапно.Очень нужно

Answer 1

$\int \frac{e^{3x}+2e^{x}}{e^{2x}+e^{x}+1}\, dx=[\; t=e^{x}\; ,\; x=lnt\; ,\; dx=\frac{dt}{t}\; ]=\int \frac{t^3+2t}{t\cdot (t^2+t+1)}\, dt=\\\\=\int \frac{t^3+2t}{t^3+t^2+t}\, dt=\int (1+\frac{2t-t^2-t}{t(t^2+t+1)})\, dt=\int (1+\frac{t-t^2}{t\, (t^2+t+1)})\, dt=\int (1-\frac{t-1}{t^2+t+1})\, dt=\\\\=t-\frac{1}{2}ln|(t+\frac{1}{2})^2+\frac{3}{4}|+\frac{3}{2}\cdot \frac{2}{\sqrt3}\cdot arctg\frac{2t+1}{\sqrt3}+C=\\\\=t-\frac{1}{2}\, ln|t^2+t+1|+\sqrt3\cdot arctg\frac{2t+1}{\sqrt3}+C=\\\\=e^{x}-\frac{1}{2}\, ln|e^{2x}+e^{x}+1|+\sqrt3\cdot arctg\frac{2e^{x}+1}{\sqrt3}+C\; ;$

$2)\; \; \int (x^2-3x)sin5x\, dx=[\; u=x^2-3x\; ,\; du=2x-3\; ,\; dv=sin5x\, dx\; ,\\\\v=-\frac{1}{5}cos5x\; ]=-\frac{x^2-3}{5}cos5x+\frac{1}{5}\int (\underbrace {2x-3}_{u})\cdot \underbrace {cos5x\, dx}_{dv}=\\\\=[\; du=2dx,v=\frac{1}{5}sin5x\; ]=-\frac{x^2-3}{5}\, cos5x+\frac{1}{5}\cdot (\frac{2x-3}{5}sin5x-\frac{2}{5}\int sin5x\, dx)=\\\\=-\frac{x^2-3}{5}\, cos5x+\frac{2x-3}{25}\, sin5x-\frac{2}{25}\cdot (-\frac{1}{5}\, cos5x)+C =$

$=-\frac{x^2-3}{5}\, cos5x+\frac{2x-3}{25}\, sin5x+\frac{2}{125}\, cos5x+C$