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

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

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integral of t/(1+t)

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

1 + t - ln ∣ 1 + t ∣ + C

Solution steps

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

Apply u-substitution

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

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

Expand

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

= \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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= u - ln ∣ u ∣

Substitute back

u = 1 + t

= 1 + t - ln ∣ 1 + t ∣

Add a constant to the solution

= 1 + t - ln ∣ 1 + t ∣ + C

6 x^{2} + 3 y^{2} = 12

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\frac{d}{d u} (u^{2} v)

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81 m^{8} + 2 m^{4} + 1