Question

Решите матешу,то что на скрине пж

Answer 1

$1.\\1)\;\int x^7dx=\frac{x^8}8+C\\\\2)\;\int\frac{dx}{x^5}=-\frac1{4x^4}+C\\\\3)\;\int(x^4-4x^3+2x)dx=\frac{x^5}5-x^4+x^2+C\\\\4)\;\int\frac{1-6x+4x^2}{x^2}dx=\int\left(\frac1{x^2}-\frac6x+4\right)dx=-\frac1{x}-6\ln x+4x+C\\\\5)\;\int\frac{x^3+8}{x^2-2x+4}dx=\frac{(x+2)(x^2-2x+4)}{x^2-2x+4}dx=\int(x+2)dx=\frac{x^2}2+2x+C\\\\6)\;\int7^xdx=\frac{7^x}{\ln7}+C\\\\7)\;8\cos xdx=8\sin x+C\\\\8)\;\int\frac{\sin x}9dx=-\frac{\cos x}9+C$

$9)\;\int\frac{x^2-7}{9-x^2}dx=-\int\frac{x^2-9+2}{x^2-9}dx=-\int\left(\frac{x^2-9}{x^2-9}+\frac2{x^2-9}\right)dx=-\int\left(1+\frac2{(x+3)(x-3)}\right)dx=\\=-\int\left(1-\frac1{3(x+3)}+\frac1{3(a-b)}\right)dx=-\left(x-\frac13\ln(x+3)+\frac13\ln(x-3)+C\right)=\\=-x+\frac13\ln(x+3)-\frac13\ln(x-3)-C$

$2.\\1)\;\int\cos5xdx=\left(\begin{array}{c}5x=t\\dx=\frac{dt}5\end{array}\right)=\int\frac{\cos tdt}5=\frac{\sin t}5+C=\frac15\sin5x+C\\\\2)\;\int(12x-5)^7dx=\left(\begin{array}{c}12x-5=t\\dx=\frac{dt}{12}\end{array}\right)=\int\frac{t^7dt}{12}=\frac{t^8}{8\cdot12}+C=\frac{(12x-5)^8}{96}+C$