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

The integral of 1/(x^2+4x+b) is 1/(sqrt(-4+b))arctan((x+2)/(sqrt(b-4)))+C

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

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

\frac{1}{- 4 + b} arctan (\frac{x + 2}{b - 4}) + C

Solution steps

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

Complete the square

x^{2} + 4 x + b : (x + 2)^{2} + b - 4

= \int \frac{1}{( x + 2 ) ^{2} + b - 4} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + b - 4} d u

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

Apply Integral Substitution

= \int \frac{1}{( v ^{2} + 1 ) - 4 + b} d v

Take the constant out:

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

= \frac{1}{- 4 + b} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{- 4 + b} arctan (v)

Substitute back

= \frac{1}{- 4 + b} arctan (\frac{x + 2}{b - 4})

Add a constant to the solution

= \frac{1}{- 4 + b} arctan (\frac{x + 2}{b - 4}) + C

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