What is the integral of 8/(x+1) ?

The integral of 8/(x+1) is 8ln|x+1|+C

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integral of 8/(x+1)

\int \frac{8}{x + 1} d x

8 ln ∣ x + 1 ∣ + C

Solution steps

\int \frac{8}{x + 1} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{x + 1} d x

Apply u-substitution

: \int \frac{1}{u} d u

= 8 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 8 ln ∣ u ∣

Substitute back

u = x + 1

= 8 ln ∣ x + 1 ∣

Add a constant to the solution

= 8 ln ∣ x + 1 ∣ + C

\frac{\partial}{\partial x} (\frac{5}{5 x + 3 y})

\frac{10}{s ( s + 2 ) ( s + 3 ) ^{2}}

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

x y^{^{'}} + y = - 5 x y^{2}, y (1) = 7

\int arccot (2 y) d y