1. Найти интеграл методом почленного интегрирования

Question 1

Question 2

$\int {\frac {x^{\frac{1}{3}}x^3}{6}-\frac {sin 3x}{2}+\frac{5}{x}} \, dx = \int {\frac {x^{\frac{1}{3}}x^3}{6} \,dx- \int \frac {sin 3x}{2} \, dx +\int \frac{5}{x}} \, dx=$

$\frac {1}{6}\int {{x^{\frac{10}{3}} \,dx-\frac{1}{2} \int {sin 3x} \, dx+ 5\int \frac{1}{x}} \, dx=$

$\frac{1}{6}\frac {x^\frac{13}{3}}{\frac{13}{3}}-\frac{1}{2} \frac{1}{3}cos (3x)+5ln|x|+C= \frac{1}{26}{x^\frac{13}{3}}-\frac{1}{6}cos (3x)+5ln|x|+C$ , c є R

$\int{\frac {6x^2+x^4-x^{\frac{1}{2}}}{x}}\, dx=\int{6x+x^3-x^{\frac{-1}{2}}\, dx= \int{6x\, dx+\int x^3\, dx-\int x^{\frac{-1}{2}}\, dx=$

$6\int{x\, dx+\int x^3\, dx-\int x^{\frac{-1}{2}}\, dx= 6\frac {x^2}{2} +\frac{x^4}{4}-\frac{x^\frac{1}{2}}{\frac{1}{2}}+c=$

$3x^2 +\frac{x^4}{4}-2x^\frac{1}{2}}+c$ ,c є R

$\int{\frac{2x}{x^3+x}-3e^x+\frac{5x^2}{3x^3}}\, dx= \int{\frac{2}{x^2+1}-3e^x+\frac{5}{3x}}\, dx= \int{\frac{2}{x^2+1}\, dx-\int3e^x\, dx+\int\frac{5}{3x}}\, dx= 2\int{\frac{1}{x^2+1}\, dx-3\inte^x\, dx+\frac{5}{3}\int\frac{1}{x}}\, dx= 2arctg(x) -3e^x+\frac{5}{3}ln|x|+c$ ,c є R