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

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

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

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

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

Solution steps

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

Complete the square

x^{2} + 4 x + 1 : (x + 2)^{2} - 3

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

Apply u-substitution

: \int \frac{u}{u ^{2} - 3} d u

= \int \frac{u}{u ^{2} - 3} d u

Apply u-substitution

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

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

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}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

= \frac{1}{2} ln (x + 2)^{2} - 3

Simplify

\frac{1}{2} ln (x + 2)^{2} - 3 : \frac{1}{2} ln x^{2} + 4 x + 1

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

Add a constant to the solution

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

\int (sinh (x) - cosh (x)) e^{- x} d x

x^{2} d x - y (1 + x^{3}) d y = 0

x \to 0 lim (\frac{x ^{5} + 3 x}{x})

\int z (a + b z^{3})^{2} d z

x \to 0 lim (\frac{1}{x cot ( x )})

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