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

The integral of (2t)/(1+t) is 2(1+t-ln|1+t|)+C

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integral of (2t)/(1+t)

\int \frac{2 t}{1 + t} d t

2 (1 + t - ln ∣ 1 + t ∣) + C

Solution steps

\int \frac{2 t}{1 + t} d t

Take the constant out:

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

= 2 \cdot \int \frac{t}{1 + t} d t

Apply u-substitution

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

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

Expand

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

= 2 \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

= 2 (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 2 (u - ln ∣ u ∣)

Substitute back

u = 1 + t

= 2 (1 + t - ln ∣ 1 + t ∣)

Add a constant to the solution

= 2 (1 + t - ln ∣ 1 + t ∣) + C

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What is the integral of (2t)/(1+t) ?
The integral of (2t)/(1+t) is 2(1+t-ln|1+t|)+C