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

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

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integral of t/(t-1)

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

t - 1 + ln ∣ t - 1 ∣ + C

Solution steps

\int \frac{t}{t - 1} 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 = t - 1

= t - 1 + ln ∣ t - 1 ∣

Add a constant to the solution

= t - 1 + ln ∣ t - 1 ∣ + C

\frac{d}{d x} (csc (5 x))

y = 3 lo g_{7} (x^{2} + 1)

\frac{d y}{d x}, 7 x^{2} - 280 x + y^{2} + 174 = 0

\frac{d}{d x} (sin (x) e^{c o s (x)})

3 t^{4}