Question

Интеграл на фото! Знатоки помогите

Answer 1

Ответ:

$\int \dfrac{1-\sqrt[3]{x+2}}{1+8\sqrt{x+2}}\, dx=\Big[\ t^6=x+2\ ,\ x=t^6-2\ ,\ dx=6t^5\, dt\ \ ,\ t=\sqrt[6]{x+2}\ \Big]=\\\\\\=\int \dfrac{1-t^2}{1+8t^3}\cdot 6t^5\, dt=-\dfrac{6}{8}\int \dfrac{t^7-t^5}{t^3+\frac{1}{8}}\, dt=-\dfrac{3}{4}\int \Bif\Big(t^4-t^2-\dfrac{1}{8}t+\dfrac{1}{64}\cdot \dfrac{8t^2+t}{t^3+\frac{1}{8}}\Big)\, dt=\\\\\\=-\dfrac{3}{4}\cdot \Big(\dfrac{t^5}{5}-\dfrac{t^3}{3}-\dfrac{t^2}{16}+\dfrac{1}{64}\int \dfrac{8t^2+t}{(t+\frac{1}{2})(t^2-\frac{1}{2}\, t+\frac{1}{8})}\Big)=I$

$\dfrac{8t^2+t}{(t+\frac{1}{2})(t^2-\frac{1}{2}\, t+\frac{1}{8})}=\dfrac{A}{t+\frac{1}{2}}}+\dfrac{Bt+C}{t^2-\frac{1}{2}\, t+\frac{1}{8}}\ ;\\\\\\8t^2+t=At^2-\dfrac{A}{2}t+\dfrac{A}{8}+Bt^2+\dfrac{B}{2}\, t +Ct+\dfrac{C}{2}\\\\\\t=-\dfrac{1}{2}\ \ \to \ \ \ A=\dfrac{8\cdot \frac{1}{4}-\frac{1}{2}}{\frac{1}{4}+\frac{1}{4}+\frac{1}{8}}=\dfrac{12}{5}=2,4\\\\\\t^2\ |\ A+B=8\ \ \ ,\ \ \ \ B=8-A=8-\dfrac{12}{5}=\dfrac{28}{5}=5,6\\t^1\ |\ -\dfrac{A}{2}+\dfrac{B}{2}+C=1\ \ \ ,\ \\t^0\ |\ \dfrac{A}{8}+\dfrac{C}{2}=0$

$-A+B+2C=2\ \ ,\ \ \ 2C=2+A-B=2+2,4-5,6=-1,2\ ,\ C=-0,6\\\\\\\int \dfrac{8t^2+t}{(t+\frac{1}{2})(t^2-\frac{1}{2}\, t+\frac{1}{8})}\, dt=\int \dfrac{2,4\, dt}{t+\frac{1}{2}}}+\int \dfrac{5,6\, t-0,6}{t^2-\frac{1}{2}\, t+\dfrac{1}{8}}\, dt=\\\\\\=2,4\cdot ln|t+0,5|+\int \dfrac{5,6\, t-0,6}{(t-\frac{1}{4})^2-\frac{1}{16}}\, dt=\Big[\ u=t-\dfrac{1}{4}\ ,\ du=dt\ \Big]=\\\\\\=2,4\cdot ln|t+0,5|+\int \dfrac{5,6(u+0,25)-0,6}{u^2-\frac{1}{16}}\, dt=$

$=2,4\cdot ln|t+0,5|+\int \dfrac{5,6u+0,8}{u^2-\frac{1}{16}}\, du=\\\\\\=2,4\cdot ln|t+0,5|+\dfrac{5,6}{2}\int \dfrac{2u\, du}{u^2-\frac{1}{16}}+0,8\int \dfrac{du}{u^2-\frac{1}{16}}=\\\\\\=2,4\cdot ln|t+0,5|+2,8\cdot ln\Big|\, u^2-\dfrac{1}{16}\, \Big|+0,8\cdot \dfrac{1}{2\cdot \frac{1}{4}}\cdot ln\Big|\dfrac{u-\frac{1}{4}}{u+\frac{1}{4}}\Big|+C=$

$=2,4\cdot ln\Big|\sqrt[6]{x+2}+0,5\Big|+2,8\cdot ln\Big|\sqrt[3]{x+2}-0,5\sqrt[6]{x+2}+0,125\Big|+\\\\\\+1,6\cdot ln\Big|\dfrac{4\sqrt[6]{x+2}-1}{4\sqrt[6]{x+2}+1}\Big|+C$

$\star \ \ \int \dfrac{1-\sqrt[3]{x+2}}{1+8\sqrt{x+2}}\, dx=-\dfrac{3}{20}\sqrt[6]{(x+2)^5}+\dfrac{1}{4}\sqrt{x+2}+\dfrac{3}{64}\sqrt[3]{x+2}-\\\\\\-\dfrac{3}{256}\cdot \Big(2,4\cdot ln\Big|\sqrt[6]{x+2}+0,5\Big|+2,8\cdot ln\Big|\sqrt[3]{x+2}-0,5\sqrt[6]{x+2}+0,125\Big|+\\\\\\+1,6\cdot ln\Big|\dfrac{4\sqrt[6]{x+2}-1}{4\sqrt[6]{x+2}+1}\Big|+C\Big)$

$P.S.$

${}\ \ \ \ t^7-t^5\ \ \ |\ \ t^3+\frac{1}{8}\\{}\qquad \qquad \ \ \ ---------\\-t^7+\frac{1}{8}\, t^4\ \ \ \ t^4-t^2-\frac{1}{8}\, t\\{}\ -----\\-t^5-\frac{1}{8}\, t^4\\-t^5-\frac{1}{8}\, t^2\\------\\-\frac{1}{8}t^4+\frac{1}{8}t^2\\-\frac{1}{8}t^2-\frac{1}{64}t\\------\\{}\ \ \ \frac{1}{8}t^2+\frac{1}{64}t$