Question

найти интеграл (3-x)dx/sqrt(1+6x-7x²)

Answer 1

$\int \dfrac{3-x}{\sqrt{1+6x-7x^2}}\, dx=-\dfrac{1}{7}\int \dfrac{(3-x)}{\sqrt{(x-\frac{3}{7})^2-\frac{16}{49}}}\, dx=\Big[\; t=x-\frac{3}{7},\; dx=dt\; \Big]=\\\\\\=-\dfrac{1}{7}\int \dfrac{\frac{18}{7}-t}{\sqrt{t^2-\frac{16}{49}}}\, dt=-\dfrac{1}{7}\cdot \Big(\dfrac{18}{7}\cdot ln\Big|t+\sqrt{t^2-\frac{16}{49}}\, \Big|-\frac{1}{2}\cdot 2\sqrt{t^2-\frac{16}{49}}\Big)+C=$

$=-\frac{18}{49}\cdot ln\, \Big|\, x-\frac{3}{7}+\sqrt{x^2-\frac{6}{7}x-\frac{1}{7}}\, \Big|+\frac{1}{7}\cdot \sqrt{x^2-\frac{6}{7}x-\frac{1}{7}}+C$