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

The integral of (sqrt(t-1))/t is 2(-arctan(sqrt(t-1))+sqrt(t-1))+C

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integral of (sqrt(t-1))/t

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

2 (- arctan (t - 1) + t - 1) + C

Solution steps

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

Apply u-substitution

= \int \frac{2 u ^{2}}{u ^{2} + 1} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2}}{u ^{2} + 1} d u

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

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

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

\int 1 d u = u

= 2 (- arctan (u) + u)

Substitute back

u = t - 1

= 2 (- arctan (t - 1) + t - 1)

Add a constant to the solution

= 2 (- arctan (t - 1) + t - 1) + C

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