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

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

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

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

ln z^{2} + 1 + z + C

Solution steps

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

Use the common integral:

\int \frac{1}{1 + z ^{2}} d z = ln z^{2} + 1 + z

= ln z^{2} + 1 + z

Add a constant to the solution

= ln z^{2} + 1 + z + C

f (x) = 4 (4 e^{x} - 25)^{7}

y = 15 - x^{2}, (1, 14)

\int x^{5} (2 - 5 x^{3})^{\frac{2}{3}} d x

y^{^{''}} + 1 = 0

\frac{d}{d y} (3 y)

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