What is the integral of (4+8xsin(x))/(2x) ?

The integral of (4+8xsin(x))/(2x) is 2(ln|x|-2cos(x))+C

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integral of (4+8xsin(x))/(2x)

\int \frac{4 + 8 x sin ( x )}{2 x} d x

2 (ln ∣ x ∣ - 2 cos (x)) + C

Solution steps

\int \frac{4 + 8 x sin ( x )}{2 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{4 + 8 x sin ( x )}{x} d x

Expand

\frac{4 + 8 x sin ( x )}{x} : \frac{4}{x} + 8 sin (x)

= \frac{1}{2} \cdot \int \frac{4}{x} + 8 sin (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

= \frac{1}{2} (\int \frac{4}{x} d x + \int 8 sin (x) d x)

\int \frac{4}{x} d x = 4 ln ∣ x ∣

\int 8 sin (x) d x = - 8 cos (x)

= \frac{1}{2} (4 ln ∣ x ∣ - 8 cos (x))

Simplify

\frac{1}{2} (4 ln ∣ x ∣ - 8 cos (x)) : 2 (ln ∣ x ∣ - 2 cos (x))

= 2 (ln ∣ x ∣ - 2 cos (x))

Add a constant to the solution

= 2 (ln ∣ x ∣ - 2 cos (x)) + C

\int tan^{8} (x) cos^{9} (x) d x

y = 1 0^{x}

(\frac{1}{3}) (\frac{d x}{d t}) - 2 e^{- t} = - x

\int_{0}^{1} x (1 - x)^{2} d x

x \to 0 lim (x (\frac{1}{x}))

What is the integral of (4+8xsin(x))/(2x) ?
The integral of (4+8xsin(x))/(2x) is 2(ln|x|-2cos(x))+C