Question

Неопределëнный интеграл ​

Answer 1

А)

$\int {sin(5-6x)} \, dx =\left[\begin{array}{cc}t=5-6x\\ & dx=-\frac{dt}{6} \\ x=\frac{5-t}{6} \end{array}\right] = -\frac{1}{6} \int {sin(t)} \, dt=\\\\\\=\frac{1}{6}\cdot cos(t)+C=\frac{1}{6}\cdot cos(5-6x)+C$

Проверка:

$\bigg[ \frac{1}{6} \cdot cos(5-6x)+C\bigg]'=\frac{1}{6} \cdot (-sin(5-6x))\cdot(-6) = sin(5-6x)$

Б)

$\int {\frac{lnx}{x} } \, dx = \int {lnx\cdot(\frac{dx}{x} })=\int {lnx} \, d(lnx) =\bigg[t=lnx \bigg]=\int {t } \, dt =\\\\=\frac{t^2}{2}+C=\frac{(lnx)^2}{2}+C$

Проверка:

$\bigg[ \frac{(lnx)^2}{2}+C \bigg]'=\frac{1}{2} \cdot 2\cdot lnx\cdot \frac{1}{x} =\frac{lnx}{x}$

В)

$\int {e^{3-4x}} \, dx =\left[\begin{array}{cc}t=3-4x\\&dx=-\frac{dt}{4} \\x=\frac{3-t}{4} \end{array}\right] =-\frac{1}{4}\int {e^{t}} \, dt=\\\\\\=-\frac{1}{4}\cdot e^t+C =-\frac{1}{4}\cdot e^{3-4x}+C$

Проверка:

$\bigg[-\frac{1}{4}\cdot e^{3-4x}+C\bigg]'=-\frac{1}{4}\cdot e^{3-4x}\cdot (-4)=e^{3-4x}$

Г)

$\int {\frac{dx}{(7x+3)^2} } =\left[\begin{array}{cc}t=7x+3\\&dx=\frac{dt}{7} \\x=\frac{t-3}{7} \end{array}\right] =\frac{1}{7} \int {\frac{dt}{t^2} } =\\\\\\=-\frac{1}{7\cdot t} +C=-\frac{1}{7\cdot (7x+3)} +C$

Проверка:

$\bigg[ -\frac{1}{7\cdot (7x+3)} +C\bigg]'=-\frac{1}{7\cdot (7x+3)^2}\cdot(-1)\cdot7 = \frac{1}{(7x+3)^2}$