What is the integral of 8-(160)/(32-x) ?

The integral of 8-(160)/(32-x) is 8x+160ln|32-x|+C

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integral of 8-(160)/(32-x)

\int 8 - \frac{160}{32 - x} d x

8 x + 160 ln ∣ 32 - x ∣ + C

Solution steps

\int 8 - \frac{160}{32 - 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 8 d x - \int \frac{160}{32 - x} d x

\int 8 d x = 8 x

\int \frac{160}{32 - x} d x = - 160 ln ∣ 32 - x ∣

= 8 x - (- 160 ln ∣ 32 - x ∣)

Simplify

= 8 x + 160 ln ∣ 32 - x ∣

Add a constant to the solution

= 8 x + 160 ln ∣ 32 - x ∣ + C

x \to 6 - lim (ln (6 - x))

- x^{2} + 2 x, 1

(x^{2} + 2 x y - y^{2}) d x + (y^{2} + 2 x y - x^{2}) d y = 0

\int x sin (\frac{x}{2}) d x

\frac{d}{d x} (4 - 2 x)

What is the integral of 8-(160)/(32-x) ?
The integral of 8-(160)/(32-x) is 8x+160ln|32-x|+C