Question

Отдаю все балы помогите прошу..

Answer 1

Неопределённый интеграл

$\displaystyle\int \dfrac{3x}{x^2-12x-8}\ dx=3\cdot \int \dfrac{x}{x^2-12x-8}\ dx=\\\\\\=3\cdot \int \dfrac{x-6+6}{x^2-12x-8}\ dx=\\\\\\=3\cdot \int \dfrac{x-6}{x^2-12x-8}\ dx+3\cdot \int \dfrac{6}{x^2-12x-8}\ dx=\\\\\\=\dfrac 32\cdot \int \dfrac{d(x^2-12x)}{x^2-12x-8}+18\cdot \int \dfrac{dx}{\big(x-6\big)^2-44}=\\\\\\=\dfrac 32\cdot \int \dfrac{d(x^2-12x-8)}{x^2-12x-8}-18\cdot \int \dfrac{d(x-6)}{\sqrt{44}^2-\big(x-6\big)^2}=$

$=\dfrac 32\cdot \ln\Big|x^2-12x-8\Big|-18\cdot \dfrac1{2\cdot\sqrt{44}}\cdot \ln\Bigg|\dfrac{\sqrt{44}+(x-6)}{\sqrt{44}-(x-6)}\Bigg|+C=\\\\\\\boldsymbol{=\dfrac 32\cdot \ln\Big|x^2-12x-8\Big|-\dfrac{9}{2\sqrt{11}}\cdot \ln\Bigg|\dfrac{2\sqrt{11}+x-6}{2\sqrt{11}-x+6}\Bigg|+C}$

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Использованы табличные интегралы

$\displaystyle\int\dfrac{dx}x=\ln|x|+C\\\\\int\dfrac{dx}{a^2-x^2}=\dfrac1{2a}\cdot \ln\Bigg|\dfrac{a+x}{a-x}\Bigg|+C\ ,\ |x|\neq |a|$