What is the integral of ((z+1))/(zsqrt(z-1)) ?

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

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integral of ((z+1))/(zsqrt(z-1))

\int \frac{( z + 1 )}{z z - 1} d z

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

Solution steps

\int \frac{z + 1}{z z - 1} d z

Apply u-substitution

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

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

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

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

Add a constant to the solution

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

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