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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

\frac{d}{d x} (8^{x} - 1 4^{x})

\frac{d y}{d t} = 2 (t + 1) y

\int_{- 1}^{3} 1 + (2 y + 2)^{2} d y

x^{3}, at x = - 2

\frac{s - 7}{s + 5}

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