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

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

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

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

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

Solution steps

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

Take the constant out:

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

= 4 \cdot \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

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

Apply u-substitution

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

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

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

= 4 \cdot \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

= 4 \cdot \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)

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

Substitute back

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

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

Simplify

4 \cdot \frac{1}{4} arctan (x - \frac{1}{2}) : arctan (x - \frac{1}{2})

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

Add a constant to the solution

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

x \to 0 lim (2 x + x)

\frac{8}{r ^{3}}

\int e^{- 2 θ} cos (θ) d θ

x \to 0 lim (x cos (\frac{2}{x}))

x \to 1 + lim (\frac{8}{1 - x})

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