What is the integral of ((z+4))/((z^2+2z+5)) ?

The integral of ((z+4))/((z^2+2z+5)) is 1/2 ln|z^2+2z+5|+3/2 arctan(1/2 (z+1))+C

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integral of ((z+4))/((z^2+2z+5))

integral

\frac{( z + 4 )}{( z ^{2} + 2 z + 5 )}

\frac{1}{2} ln z^{2} + 2 z + 5 + \frac{3}{2} arctan (\frac{1}{2} (z + 1)) + C

Solution steps

\int \frac{z + 4}{z ^{2} + 2 z + 5} d z

Complete the square

z^{2} + 2 z + 5 : (z + 1)^{2} + 4

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

Apply u-substitution

= \int \frac{u + 3}{u ^{2} + 4} d u

Expand

\frac{u + 3}{u ^{2} + 4} : \frac{u}{u ^{2} + 4} + \frac{3}{u ^{2} + 4}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{u}{u ^{2} + 4} d u + \int \frac{3}{u ^{2} + 4} d u

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

\int \frac{3}{u ^{2} + 4} d u = \frac{3}{2} arctan (\frac{u}{2})

= \frac{1}{2} ln u^{2} + 4 + \frac{3}{2} arctan (\frac{u}{2})

Substitute back

u = z + 1

= \frac{1}{2} ln (z + 1)^{2} + 4 + \frac{3}{2} arctan (\frac{z + 1}{2})

Simplify

\frac{1}{2} ln (z + 1)^{2} + 4 + \frac{3}{2} arctan (\frac{z + 1}{2}) : \frac{1}{2} ln z^{2} + 2 z + 5 + \frac{3}{2} arctan (\frac{1}{2} (z + 1))

= \frac{1}{2} ln z^{2} + 2 z + 5 + \frac{3}{2} arctan (\frac{1}{2} (z + 1))

Add a constant to the solution

= \frac{1}{2} ln z^{2} + 2 z + 5 + \frac{3}{2} arctan (\frac{1}{2} (z + 1)) + C

What is the integral of ((z+4))/((z^2+2z+5)) ?
The integral of ((z+4))/((z^2+2z+5)) is 1/2 ln|z^2+2z+5|+3/2 arctan(1/2 (z+1))+C