What is the integral of 1/(z^2+6z+10) ?

The integral of 1/(z^2+6z+10) is arctan(z+3)+C

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integral of 1/(z^2+6z+10)

\int \frac{1}{z ^{2} + 6 z + 10} d z

arctan (z + 3) + C

Solution steps

\int \frac{1}{z ^{2} + 6 z + 10} d z

Complete the square

z^{2} + 6 z + 10 : (z + 3)^{2} + 1

= \int \frac{1}{( z + 3 ) ^{2} + 1} d z

Apply u-substitution

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

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

Use the common integral:

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

= arctan (u)

Substitute back

u = z + 3

= arctan (z + 3)

Add a constant to the solution

= arctan (z + 3) + C

\frac{d}{d x} (3 x^{2} + 7 x - 2)

\frac{d x}{d y} = x 1 - y^{2}

\frac{x}{3}

y^{^{''}} - 7 y^{^{'}} + 10 y = e^{3 x}

\frac{\partial}{\partial x} (2 \cdot z \cdot cos (x))

What is the integral of 1/(z^2+6z+10) ?
The integral of 1/(z^2+6z+10) is arctan(z+3)+C