What is the integral of sqrt(4x^2-8x+5) ?

The integral of sqrt(4x^2-8x+5) is 1/4 (2(x-1)sqrt(4x^2-8x+5)+ln|2(x-1)+sqrt(4x^2-8x+5)|)+C

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integral of sqrt(4x^2-8x+5)

\int 4 x^{2} - 8 x + 5 d x

\frac{1}{4} (2 (x - 1) 4 x^{2} - 8 x + 5 + ln 2 (x - 1) + 4 x^{2} - 8 x + 5) + C

Solution steps

\int 4 x^{2} - 8 x + 5 d x

Complete the square

4 x^{2} - 8 x + 5 : 4 (x - 1)^{2} + 1

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

Apply u-substitution

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

Apply Trigonometric Substitution

= \int \frac{1}{2} sec^{3} (v) d v

Take the constant out:

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

= \frac{1}{2} \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= \frac{1}{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) ∣

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

Substitute back

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

Simplify

\frac{1}{2} (\frac{sec ^{2} ( arctan ( 2 ( x - 1 ) ) ) sin ( arctan ( 2 ( x - 1 ) ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 (x - 1))) + sec (arctan (2 (x - 1))) ∣) : \frac{1}{4} (2 (x - 1) 4 x^{2} - 8 x + 5 + ln 2 (x - 1) + 4 x^{2} - 8 x + 5)

= \frac{1}{4} (2 (x - 1) 4 x^{2} - 8 x + 5 + ln 2 (x - 1) + 4 x^{2} - 8 x + 5)

Add a constant to the solution

= \frac{1}{4} (2 (x - 1) 4 x^{2} - 8 x + 5 + ln 2 (x - 1) + 4 x^{2} - 8 x + 5) + C

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What is the integral of sqrt(4x^2-8x+5) ?
The integral of sqrt(4x^2-8x+5) is 1/4 (2(x-1)sqrt(4x^2-8x+5)+ln|2(x-1)+sqrt(4x^2-8x+5)|)+C