Question

Пожалуйстаааа, вычислить неопределенные интегралы методом замены переменной

Answer 1

$\displaystyle \int\frac{x^2dx}{\sqrt{x^2+7}}=-7\int\frac{1}{(t^2-1)^2}dt=\\=-7\int(-\frac{1}{4(t-1)}+\frac{1}{4(t-1)^2}+\frac{1}{4(t+1)}+\frac{1}{4(t+1)^2})dt=\\=\frac{7}{4}ln|t-1|+\frac{7}{4(t-1)}-\frac{7}{4}ln|t+1|+\frac{7}{4(t+1)}+C=\\=\frac{7}{4}ln|\frac{t-1}{t+1}|+\frac{7t}{2(t^2-1)}+C=\\=\frac{7}{4}ln|\frac{\sqrt{x^2+7}-x}{\sqrt{x^2+7}+x}|+\frac{x}{2}\sqrt{x^2+7}+C$

$\displaystyle \1+\frac{7}{x^2}=t^2\to x^2=\frac{7}{t^2-1}\\\sqrt{x^2+7}=\sqrt{\frac{7t^2}{t^2-1}}=\frac{t\sqrt7}{\sqrt{t^2-1}}\\x^2dx=-\frac{7t\sqrt7}{(t^2-1)^2\sqrt{t^2-1}}dt\\\frac{1}{(t^2-1)^2}=\frac{A}{t-1}+\frac{B}{(t-1)^2}+\frac{C}{t+1}+\frac{D}{(t+1)^2}=\\=-\frac{1}{4(t-1)}+\frac{1}{4(t-1)^2}+\frac{1}{4(t+1)}+\frac{1}{4(t+1)^2}\\1=A(t^3+t^2-t-1)+B(t^2+2t+1)+C(t^3-t^2-t+1)+D(t^2-2t+1)\\t^3|0=A+C\\t^2|0=A+B-C+D\\t|0=-A+2B-C-2D\\t^0|1=-A+B+C+D\\A=-\frac{1}{4};B=\frac{1}{4};C=\frac{1}{4};D=\frac{1}{4}$

$\displaystyle \int\frac{2x-3}{x^2-3x+2}dx=\int\frac{d(x^2-3x+2)}{x^2-3x+2}=ln|x^2-3x+2|+C$