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

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

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integral of 1/(4v^2-4v+5)

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

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

Solution steps

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

Complete the square

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

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

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 = v - \frac{1}{2}

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

Add a constant to the solution

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

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