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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Apply Trigonometric Substitution

= 2 \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= 2 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v)

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= 2 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 2 (\frac{sec ^{2} ( arctan ( x - \frac{1}{2} ) ) sin ( arctan ( x - \frac{1}{2} ) )}{2} + \frac{1}{2} ln tan (arctan (x - \frac{1}{2})) + sec (arctan (x - \frac{1}{2})))

Simplify

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

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

Add a constant to the solution

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

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