Question

antiderivatives: problem 1
(1 point)
calculate the following antiderivatives:
(a) $\int \frac{5}{x} d x=\square +c$
(b) $\int-4 \sin x+6 \cos x d x=\square +c$
(c) $\int-9 e^{x} d x=\square +c$
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Explanation:

Step1: Solve \(\int\frac{5}{x}dx\)

Use the formula \(\int\frac{a}{x}dx=a\ln|x| + C\) (\(a = 5\)).
So \(\int\frac{5}{x}dx=5\ln|x|+C\).

Step2: Solve \(\int(- 4\sin x + 6\cos x)dx\)

Use the formulas \(\int\sin xdx=-\cos x + C\) and \(\int\cos xdx=\sin x + C\).
\(\int(-4\sin x + 6\cos x)dx=-4\int\sin xdx+6\int\cos xdx\)
\(=-4(-\cos x)+6\sin x + C = 4\cos x+6\sin x + C\).

Step3: Solve \(\int-9e^{x}dx\)

Use the formula \(\int e^{x}dx=e^{x}+C\).
\(\int-9e^{x}dx=-9\int e^{x}dx=-9e^{x}+C\).

Answer:

(a) \(5\ln|x|\)
(b) \(4\cos x + 6\sin x\)
(c) \(-9e^{x}\)