What is the integral from 0 to t of y/(y+2) ?

The integral from 0 to t of y/(y+2) is t-2(ln|2+t|-ln(2))

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integral from 0 to t of y/(y+2)

\int_{0}^{t} \frac{y}{y + 2} d y

t - 2 (ln ∣ 2 + t ∣ - ln (2))

Solution steps

\int_{0}^{t} \frac{y}{y + 2} d y

Apply u-substitution

= \int_{2}^{2 + t} \frac{u - 2}{u} d u

Expand

\frac{u - 2}{u} : 1 - \frac{2}{u}

= \int_{2}^{2 + t} 1 - \frac{2}{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

= \int_{2}^{2 + t} 1 d u - \int_{2}^{2 + t} \frac{2}{u} d u

\int_{2}^{2 + t} 1 d u = t

\int_{2}^{2 + t} \frac{2}{u} d u = 2 (ln ∣ 2 + t ∣ - ln (2))

= t - 2 (ln ∣ 2 + t ∣ - ln (2))

What is the integral from 0 to t of y/(y+2) ?
The integral from 0 to t of y/(y+2) is t-2(ln|2+t|-ln(2))