Question

Решите матешу,то что на скрине пж

Answer 1

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

$9);intfrac{x^2-7}{9-x^2}dx=-intfrac{x^2-9+2}{x^2-9}dx=-intleft(frac{x^2-9}{x^2-9}+frac2{x^2-9}right)dx=-intleft(1+frac2{(x+3)(x-3)}right)dx=\=-intleft(1-frac1{3(x+3)}+frac1{3(a-b)}right)dx=-left(x-frac13ln(x+3)+frac13ln(x-3)+Cright)=\=-x+frac13ln(x+3)-frac13ln(x-3)-C$

$2.\1);intcos5xdx=left(begin{array}{c}5x=t\dx=frac{dt}5end{array}right)=intfrac{cos tdt}5=frac{sin t}5+C=frac15sin5x+C\\2);int(12x-5)^7dx=left(begin{array}{c}12x-5=t\dx=frac{dt}{12}end{array}right)=intfrac{t^7dt}{12}=frac{t^8}{8cdot12}+C=frac{(12x-5)^8}{96}+C$