What is the integral from 0 to x of t/(t+5) ?

The integral from 0 to x of t/(t+5) is x-5(ln|x+5|-ln(5))

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integral from 0 to x of t/(t+5)

\int_{0}^{x} \frac{t}{t + 5} d t

x - 5 (ln ∣ x + 5 ∣ - ln (5))

Solution steps

\int_{0}^{x} \frac{t}{t + 5} d t

Apply u-substitution

: \int_{5}^{x + 5} \frac{u - 5}{u} d u

= \int_{5}^{x + 5} \frac{u - 5}{u} d u

Expand

\frac{u - 5}{u} : 1 - \frac{5}{u}

= \int_{5}^{x + 5} 1 - \frac{5}{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_{5}^{x + 5} 1 d u - \int_{5}^{x + 5} \frac{5}{u} d u

\int_{5}^{x + 5} 1 d u = x

\int_{5}^{x + 5} \frac{5}{u} d u = 5 (ln ∣ x + 5 ∣ - ln (5))

= x - 5 (ln ∣ x + 5 ∣ - ln (5))

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What is the integral from 0 to x of t/(t+5) ?
The integral from 0 to x of t/(t+5) is x-5(ln|x+5|-ln(5))