What is the integral of 1/(sqrt(4z+z^2)) ?

The integral of 1/(sqrt(4z+z^2)) is ln(1/2 |sqrt(z^2+4z)+z+2|)+C

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integral of 1/(sqrt(4z+z^2))

\int \frac{1}{4 z + z ^{2}} d z

ln (\frac{1}{2} z^{2} + 4 z + z + 2) + C

Solution steps

\int \frac{1}{4 z + z ^{2}} d z

Complete the square

4 z + z^{2} : (z + 2)^{2} - 4

= \int \frac{1}{( z + 2 ) ^{2} - 4} d z

Apply u-substitution

= \int \frac{1}{u ^{2} - 4} d u

Apply Trigonometric Substitution

= \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln tan (arcsec (\frac{1}{2} (z + 2))) + sec (arcsec (\frac{1}{2} (z + 2)))

Simplify

ln tan (arcsec (\frac{1}{2} (z + 2))) + sec (arcsec (\frac{1}{2} (z + 2))) : ln (\frac{1}{2} z^{2} + 4 z + z + 2)

= ln (\frac{1}{2} z^{2} + 4 z + z + 2)

Add a constant to the solution

= ln (\frac{1}{2} z^{2} + 4 z + z + 2) + C

What is the integral of 1/(sqrt(4z+z^2)) ?
The integral of 1/(sqrt(4z+z^2)) is ln(1/2 |sqrt(z^2+4z)+z+2|)+C