Question

Нужно решить интегралы Под цифрой 9

Answer 1

$9a)\ \displaystyle \int \frac{x^2\, dx}{5-x^6}=\frac{1}{3}\int \frac{3x^2\, dx}{5-(x^3)^2}=\frac{1}{3}\int \frac{d(x^3)}{5-(x^3)^2}=\frac{1}{3}\cdot \frac{1}{2\sqrt5}\cdot ln\Big|\, \frac{\sqrt5+x}{\sqrt5-x}\, \Big|+C\\\\\\\\b)\ \ \int \frac{dx}{\sqrt{x(3x+5)}}=\int \frac{dx}{\sqrt{3x^2+5x}}=\frac{1}{\sqrt3}\int \frac{dx}{\sqrt{x^2+\frac{5}{3}\, x}}=$

$\displaystyle =\frac{1}{\sqrt3}\int \frac{dx}{\sqrt{(x+\frac{5}{6})^2-\frac{25}{36}}}=\frac{1}{\sqrt3}\cdot ln\Big|\ x+\frac{5}{6}+\sqrt{\Big(x+\frac{5}{6}\Big)^2-\frac{25}{36}}\ \Big|+C=\\\\\\=\frac{1}{\sqrt3}\cdot ln\Big|\ x+\frac{5}{6}+\sqrt{x^2+\frac{5}{3}\, x}\ \Big|+C$

$\displaystyle c)\ \ \int x\cdot arctg\sqrt{x^2-1}\, dx=\Big[\ t=x^2-1\ ,\ dt=2x\, dx\ \Big]=\frac{1}{2}\int arctg\sqrt{t}\, dt=\\\\\\=\Big[\ u=arctg\sqrt{t}\ ,\ du=\frac{dt}{2\sqrt{t}(1+t)}\ ,\ dv=dt\ ,\ v=t\ \Big]=uv-\int v\, du=\\\\\\=\frac{1}{2}t\cdot arctg\sqrt{t}-\frac{1}{2}\int \frac{\sqrt{t}\, dt}{2(1+t)}=\Big[\ t=z^2\ ,\ dt=2z\, dz\ \Big]=$

$\displaystyle =\frac{1}{2}t\cdot arctg\sqrt{t}-\frac{1}{2}\int \frac{2z^2\, dz}{2(1+z^2)}=\frac{1}{2}t\cdot arctg\sqrt{t}-\frac{1}{2}\int \Big(1-\frac{1}{1+z^2}\Big)\, dz=$

$\displaystyle =\frac{1}{2}t\cdot arctg\sqrt{t}-\frac{z}{2}+\frac{1}{2}arctgz+C=\\\\\\=\frac{1}{2}(x^2-1)\cdit arctg\sqrt{x^2-1}-\frac{1}{2}\sqrt{x^2-1}+\frac{1}{2}arctg\sqrt{x^2-1}+C$