Question

calculate the following antiderivatives:
(a) $$ \int \frac { 6 } { x } d x = $$ +c.
(b) $$ \int - 6 \sin x + 3 \cos x d x = $$ +c.
(c) $$ \int - 2 e ^ { x } d x = $$ +c.

Explanation:

Step1: Solve (a)

Use the formula $\int\frac{a}{x}dx=a\ln|x| + C$ ($a$ is a constant).
For $\int\frac{6}{x}dx$, here $a = 6$, so $\int\frac{6}{x}dx=6\ln|x|+C$.

Step2: Solve (b)

Use the formulas $\int\sin xdx=-\cos x + C$ and $\int\cos xdx=\sin x + C$.
For $\int(-6\sin x + 3\cos x)dx$, we have:
$\int(-6\sin x+3\cos x)dx=-6\int\sin xdx + 3\int\cos xdx$.
Substitute the formulas: $-6(-\cos x)+3\sin x + C=6\cos x+3\sin x + C$.

Step3: Solve (c)

Use the formula $\int e^{x}dx=e^{x}+C$.
For $\int - 2e^{x}dx$, we have $-2\int e^{x}dx$.
Substitute the formula: $-2e^{x}+C$.

Answer:

(a) $6\ln|x|$; (b) $6\cos x + 3\sin x$; (c) $-2e^{x}$