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

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

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

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

\frac{1}{2} ln \frac{4 x ^{2} - 4 x + 5}{2} + 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}{2 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}{2} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

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

Substitute back

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

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

Simplify

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

= \frac{1}{2} ln \frac{4 x ^{2} - 4 x + 5}{2} + x - \frac{1}{2}

Add a constant to the solution

= \frac{1}{2} ln \frac{4 x ^{2} - 4 x + 5}{2} + x - \frac{1}{2} + C

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