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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

\frac{1}{4 u ^{2} + 4} = \frac{1}{4} \frac{1}{u ^{2} + 1}

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

Take the constant out:

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

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

Use the common integral:

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

= \frac{1}{4} arctan (u)

Substitute back

u = x + \frac{1}{2}

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

Add a constant to the solution

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

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

\frac{d}{d θ} (1 + cos (2 θ))

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

(2 cos (2 x))^{^{'}}

3 s + 2

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