Question

алгебра, вычисление интегралов

Answer 1

$1) int6^{5x+2}dx=frac15int6^{5x+2}d(5x+2)=frac{1}{5}*frac{1}{ln6}*6^{5x+2}+C=frac{6^{5x+2}}{5ln6}+C \ \ 3) int xe^{-x^2}dx=-frac12 int e^{-x^2}d(-x^2)=-frac12 e^{-x^2}+C \ \ 4) int {frac{sqrt[4]{lnx^2}}{x}}dx=int {frac{sqrt[4]{2lnx}}{x}}dx=sqrt[4]2int {frac{sqrt[4]{lnx}}{x}}dx=sqrt[4]2int {sqrt[4]{lnx}}d(lnx)=sqrt[4]2int (lnx)^{frac14}d(lnx)= \ =sqrt[4]2*frac{(lnx)^{frac14+1}}{frac14+1}+C=frac{4sqrt[4]{2}}{5}ln^{frac54}x+C$

$2) int frac{x-1}{sqrt{2x-x^2}}dx=int frac{x-1}{sqrt{x(2-x)}}dx=[u=1-x; x=1-u; dx=-du]= \ \ =-int frac{1-u-1}{sqrt{(1-u)(2-(1-u))}}du=int frac{udu}{sqrt{(1-u)(1+u)}}=int frac{udu}{sqrt{1-u^2}}=-frac{1}{2}int frac{d(1-u^2)}{sqrt{1-u^2}}= \ \ =-sqrt{1-u^2}+C=-sqrt{1-(1-x)^2}+C=-sqrt{1-(1-2x+x^2)}+C= \ \ =-sqrt{2x-x^2}+C$