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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Expand

\frac{u + 1}{u ^{2} + 1} : \frac{u}{u ^{2} + 1} + \frac{1}{u ^{2} + 1}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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