What is the integral of (4y)/(y+1) ?

The integral of (4y)/(y+1) is 4(y+1-ln|y+1|)+C

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integral of (4y)/(y+1)

\int \frac{4 y}{y + 1} d y

4 (y + 1 - ln ∣ y + 1 ∣) + C

Solution steps

\int \frac{4 y}{y + 1} d y

Take the constant out:

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

= 4 \cdot \int \frac{y}{y + 1} d y

Apply u-substitution

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

= 4 \cdot \int \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= 4 \cdot \int 1 - \frac{1}{u} d u

Apply the Sum Rule:

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

= 4 (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 4 (u - ln ∣ u ∣)

Substitute back

u = y + 1

= 4 (y + 1 - ln ∣ y + 1 ∣)

Add a constant to the solution

= 4 (y + 1 - ln ∣ y + 1 ∣) + C

\int 2 sin (θ) d θ

f (x) = 2 x^{3} - 9 x^{2} - 24 x + 2

\frac{4}{3} π x^{3}

\frac{d}{d t} (e^{s i n (t)})

\frac{1 + x}{1 - x}

What is the integral of (4y)/(y+1) ?
The integral of (4y)/(y+1) is 4(y+1-ln|y+1|)+C