What is the integral of 1/(x^2+4x+5) ?

The integral of 1/(x^2+4x+5) is arctan(x+2)+C

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integral of 1/(x^2+4x+5)

\int \frac{1}{x ^{2} + 4 x + 5} d x

arctan (x + 2) + C

Solution steps

\int \frac{1}{x ^{2} + 4 x + 5} d x

Complete the square

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

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

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 = x + 2

= arctan (x + 2)

Add a constant to the solution

= arctan (x + 2) + C

f (x) = \frac{9}{x - 3}, at x = 2

y = e^{5}

\frac{d}{d x} (x^{\frac{3}{4}} - 3 x^{\frac{1}{4}})

(3.3 x + 5)^{0.37}

\frac{d}{d x} (\frac{1}{2 x ^{\frac{5}{2}}})

What is the integral of 1/(x^2+4x+5) ?
The integral of 1/(x^2+4x+5) is arctan(x+2)+C