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

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

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

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

\frac{1}{2} ln x^{2} + 4 x + 5 + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 4 x + 5

= \frac{1}{2} ln x^{2} + 4 x + 5

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + 5 + C

\frac{d}{d x} (- 4 x y)

\int tan^{3} (x) sin (x) d x

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

x \to 1 lim (\frac{x ^{2} - 2}{x - 1})

\frac{\partial}{\partial y} (\frac{y}{1 + x ^{2} + y ^{2}})

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