What is the integral of [cot(x)+1/x ] ?

The integral of [cot(x)+1/x ] is ln|sin(x)|+ln|x|+C

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integral of [cot(x)+1/x ]

\int [cot (x) + \frac{1}{x}] d x

ln ∣ sin (x) ∣ + ln ∣ x ∣ + C

Solution steps

\int cot (x) + \frac{1}{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int cot (x) d x + \int \frac{1}{x} d x

\int cot (x) d x = ln ∣ sin (x) ∣

\int \frac{1}{x} d x = ln ∣ x ∣

= ln ∣ sin (x) ∣ + ln ∣ x ∣

Add a constant to the solution

= ln ∣ sin (x) ∣ + ln ∣ x ∣ + C

\int_{0}^{1} t^{4} + t^{2} d t

\int \frac{x ^{2} + 16}{x} d x

\int_{0}^{2 π} 5 cos (x) d x

y^{^{'}} = \frac{y \cdot ln ( x )}{x \cdot ln ( y )}

n = 1 \sum \infty \frac{x ^{n}}{9 ^{n}}

What is the integral of [cot(x)+1/x ] ?
The integral of [cot(x)+1/x ] is ln|sin(x)|+ln|x|+C