What is the integral of (2y)/(x+1) ?

The integral of (2y)/(x+1) is 2yln|1+x|+C

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integral of (2y)/(x+1)

\int \frac{2 y}{x + 1} d x

2 y ln ∣ 1 + x ∣ + C

Solution steps

\int \frac{2 y}{x + 1} d x

Take the constant out:

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

= 2 y \cdot \int \frac{1}{1 + x} d x

Apply u-substitution

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

= 2 y \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 y ln ∣ u ∣

Substitute back

u = 1 + x

= 2 y ln ∣ 1 + x ∣

Add a constant to the solution

= 2 y ln ∣ 1 + x ∣ + C

x \to 0 - lim (- x + 5)

y^{^{''}} + 16 y = 16 sec (24 t)

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

(0.5) (100)

\frac{s}{s ^{2} - 1}