Ещё один Интеграл Для мозговитых людей Буду благодарен

Question

Математика

Souzhamda

5 год назад

Ещё один Интеграл Для мозговитых людей Буду благодарен

ОТВЕТЫ

Мартын

Nov 29, 2020

$\int \dfrac{dx}{x^3\sqrt{x^2+36}}=\Big[\ x=6tgt\ ,\ dx=\dfrac{6\, dt}{cos^2t}\ ,\ t=arctg\dfrac{x}{6}\ ,\\\\\\\sqrt{x^2+36}=\sqrt{36tg^2t+36}=6\sqrt{1+tg^2t}=6\cdot \sqrt{\dfrac{1}{cos^2t}}=\dfrac{6}{cost}\ \Big]=\\\\\\=\int \dfrac{\dfrac{6\, dt}{cos^2t}}{6^3\, tg^3t\cdot \dfrac{6}{cost}}=\dfrac{1}{6^3}\int \dfrac{dt}{cost\cdot tg^3t}=\dfrac{1}{216}\int \dfrac{dt}{\dfrac{sin^3t}{cos^2t}}=\dfrac{1}{216}\int \dfrac{cos^2t\, dt}{sin^3t}=$

$=\dfrac{1}{216}\int \dfrac{cost\cdot cost\, dt}{sin^3t}=\Big[\ u=cost\ ,\ du=-sint\, dt\ ,\ dv=\dfrac{cost\, dt}{sin^3t}\ ,\\\\\\v=\dfrac{(sint)^{-2}}{-2}=-\dfrac{1}{2sin^2t}\ \Big]=\dfrac{1}{216}\cdot \Big(-\dfrac{cost}{2sin^2t}-\int \dfrac{sint\, dt}{2sin^2t}\Big)=\\\\\\=\dfrac{1}{216}\cdot \Big(-\dfrac{cost}{2sin^2t}-\int \dfrac{sint\, dt}{2(1-cos^2t)}\Big)=-\dfrac{1}{432}\cdot \dfrac{cost}{sin^2t}-\dfrac{1}{432}\cdot \int \dfrac{-d(cost)}{1-cos^2t}=$

$=-\dfrac{1}{432}\cdot \dfrac{cost}{sin^2t}+\dfrac{1}{432}\cdot \dfrac{1}{2}\cdot ln\Big|\dfrac{1+cost}{1-cost}\Big|+C=\\\\\\=-\dfrac{1}{432}\cdot \dfrac{cos(arctg\frac{x}{6})}{sin^2(arctg\frac{x}{6})}+\dfrac{1}{432}\cdot \dfrac{1}{2}\cdot ln\Big|\dfrac{1+cos(arctg\frac{x}{6})}{1-cos(arctg\frac{x}{6})}\Big|+C=\\\\\\=-\dfrac{1}{432}\cdot \dfrac{\frac{6}{\sqrt{x^2+36}}}{\frac{x^2}{x^2+36}}+\dfrac{1}{432}\cdot \dfrac{1}{2}\cdot ln\Big|\dfrac{1+\frac{6}{\sqrt{x^2+36}}}{1-\frac{6}{\sqrt{x^2+36}}}\Big|+C=$

$=-\dfrac{1}{432}\cdot \dfrac{6\sqrt{x^2+36}}{x^2}+\dfrac{1}{864}\cdot ln\Big|\dfrac{\sqrt{x^2+36}+6}{\sqrt{x^2+36}-6}\Big|+C$

130