Question

пожалуйста, срочно помогите

Answer 1

Ответ:

$1)\ \ \ \displaystyle \int 2^{x}\cdot 7^{x}\, dx=\int 14^{x}\, dx=\frac{14^{x}}{ln14}+C\\\\\\2)\ \ \int \frac{(8-x)^2}{x^2}\, dx=\int \frac{64-16x+x^2}{x^2}\, dx=\int \Big(\frac{64}{x^2}-\frac{16}{x}+1\Big)\, dx=\\\\\\=-\frac{64}{x}-16\, ln|x|+x+C\\\\\\3)\ \ \int 6^{7x+1}\, dx=\int 6^{7x}\cdot 6\, dx=6\int (6^7)^{x}\, dx=6\cdot \frac{(6^7)^{x}}{ln6^7}+C=\frac{6\cdot (6^7)^{x}}{7\, ln6}+C$

$b)\ \ \displaystyle \int\limits^{ln\sqrt3}_0\, \frac{e^{x}\, dx}{1+e^{2x}}=\Big[\ t=e^{x}\ ,\ dt=e^{x}\, dx\ ,\ e^{2x}=(e^{x})^2=t^2\ \Big]=\\\\\\=\Big[\ t_1=e^0=1\ ,\ t_2=e^{ln\sqrt3}=\sqrt3\ \Big]=\int\limits^{\sqrt3}_1\, \frac{dt}{1+t^2}\ dx=arctgt\Big|_1^{\sqrt3}=\\\\\\=arctg\sqrt3-arctg1=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$

$c)\ \ \displaystyle \int\limits^1_0\, arctgx\, dx=\Big[\ u=arctgx\ ,\ du=\frac{dx}{1+x^2}\ ,\ dv=dx\ ,\ v=x\ \Big]=\\\\\\=x\cdot arctgx\Big|_0^1-\int \limits _0^1\frac{x\, dx}{1+x^2}=arctg1-\frac{1}{2}\int \limits _0^1\, \frac{2x\, dx}{1+x^2}=\Big[\ d(1+x^2)=2x\, dx\ \Big]=\\\\\\=\frac{\pi}{4}-\frac{1}{2}\cdot ln(1+x^2)\Big|_0^1=\frac{\pi}{4}-\frac{1}{2}\cdot (ln2-ln1)= \frac{\pi}{4}-\frac{ln2}{2}=\frac{\pi -2\, ln2}{4}$