Question

Проинтегрировать функции заменой переменной:

10^2x+1 dx

sin x/2 dx

dx/5x+3

Answer 1

Пошаговое объяснение:

1)

$\int\ {10^{2x+1} } \, dx =\left[\begin{array}{ccc}u=2x+1\\du=2dx\\\end{array}\right]=\frac{1}{2} \int {10^u} \, du=\frac{1}{2}*\frac{10^u}{ln10} =\frac{10^{2x+1} }{lnx} +C$

2)

$\int{sin{\frac{x}{2} } \, dx \left[\begin{array}{ccc}u=x/2\\du = \frac{dx}{2} \\\end{array}\right] = \int{sinu} \, du=-2cosu=-2cos(\frac{x}{2} )+C$

3)

$\int {\frac{1}{5x+3} } \, dx =\left[\begin{array}{ccc}u=5x+3\\du=5dx\\\end{array}\right] =\int {\frac{1}{5u} } \, du=\frac{lnu}{5} =\frac{ln(5x+3)}{5} +C$

4)

$\int {\frac{1}{x*lnx}} \, dx =\left[\begin{array}{ccc}u=lnx\\du = \frac{dx}{x} \\\end{array}\right] =\int {\frac{1}{u} } \, du=lnu = ln(lnx) +C$

5)

$\int {sin2x} \, dx =\left[\begin{array}{ccc}u=2x\\du=2dx\\\end{array}\right] =\int {\frac{1}{2} sinu} \, du =-\frac{1}{2} cosu=-\frac{1}{2} cos(2x)+C$

6)

$\int {3^{7x-1} } \, dx =\left[\begin{array}{ccc}u=7x-1\\du=7dx\\\end{array}\right] =\int{\frac{3^u}{7} } \, du=\frac{3^u}{7ln(3)} =\frac{3^{7x-1} }{7ln(3)} +C$