Question

интеграл x*3^x/2dx
интеграл по частям, прошу)

Answer 1

Ответ:

Пошаговое объяснение:

$\displaystyle \int {fg'} \, =fg-\int {f'g} \,$

$\displaystyle \int {\frac{x3^x}{2} } \, dx =\frac{1}{2} \bigg (\left[\begin{array}{ccc}f=x\quad f'=1\\g'=3^x\quad \displaystyle g=\frac{3^x}{ln(3)} \\\end{array}\right] \bigg )=\frac{1}{2} \bigg (\frac{x3^x}{ln(3)} -\frac{1}{ln(3)} \int {3^x} \, dx \bigg )=$$\displaystyle =\frac{1}{2} \bigg (\frac{x3^x}{ln(3)} -\frac{1}{ln(3)} \left[\begin{array}{ccc}\int {a^x} \, dx =\displaystyle \frac{a^x}{ln(a)},\quad for\quad a=3 \\\\\end{array}\right] \bigg )=$

$\displaystyle =\frac{1}{2} \bigg (\frac{x3^x}{ln(3)} -\frac{1}{ln(3)} *\frac{3^x}{ln(3)} \bigg )+C=\frac{3^x(xln(3)-1)}{2ln^2(3)} +C$