Question

antiderivatives: problem 1
(1 point)
calculate the following antiderivatives:
(a) $\int \frac{3}{x} d x=\square+c$.
(b) $\int-2 \sin x+9 \cos x d x=\square+c$.
(c) $\int-9 e^{x} d x=\square+c$.

Explanation:

Step1: Solve (a)

Use the formula $\int\frac{a}{x}dx=a\ln|x|+C$.
For $\int\frac{3}{x}dx$, here $a = 3$, so $\int\frac{3}{x}dx=3\ln|x|+C$.

Step2: Solve (b)

Use the formulas $\int\sin xdx=-\cos x + C$ and $\int\cos xdx=\sin x + C$.
$\int(- 2\sin x+9\cos x)dx=-2\int\sin xdx + 9\int\cos xdx$.
Substitute the formulas: $-2(-\cos x)+9\sin x + C=2\cos x+9\sin x + C$.

Step3: Solve (c)

Use the formula $\int e^{x}dx=e^{x}+C$.
For $\int - 9e^{x}dx$, since $\int ae^{x}dx=ae^{x}+C$ ($a=-9$), so $\int - 9e^{x}dx=-9e^{x}+C$.

Answer:

(a) $3\ln|x|$
(b) $2\cos x + 9\sin x$
(c) $-9e^{x}$