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

The integral of 1/(1+z^2) is arctan(z)+C

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integral of 1/(1+z^2)

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

arctan (z) + C

Solution steps

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

Use the common integral:

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

= arctan (z)

Add a constant to the solution

= arctan (z) + C

t^{2} y^{^{''}} + 13 t y^{^{'}} + 36 y = 0

9 y^{^{''}} - 60 y^{^{'}} + 51 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\frac{d}{d x} ((sin (2 x) + 1)^{2})

y = x^{2} - 1, (- 2, 3)

f (x) = 4 x^{2} - 4 x + 1, at x = 4