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

The integral of sqrt(1+(2x-4)^2) is 1/4 (2(x-2)sqrt(4x^2-16x+17)+ln|2(x-2)+sqrt(4x^2-16x+17)|)+C

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

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

\frac{1}{4} (2 (x - 2) 4 x^{2} - 16 x + 17 + ln 2 (x - 2) + 4 x^{2} - 16 x + 17) + C

Solution steps

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

Expand

1 + (2 x - 4)^{2} : 4 x^{2} - 16 x + 17

= \int 4 x^{2} - 16 x + 17 d x

Complete the square

4 x^{2} - 16 x + 17 : 4 (x - 2)^{2} + 1

= \int 4 (x - 2)^{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 - 2 ) ) ) sin ( arctan ( 2 ( x - 2 ) ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 (x - 2))) + sec (arctan (2 (x - 2))) ∣)

Simplify

\frac{1}{2} (\frac{sec ^{2} ( arctan ( 2 ( x - 2 ) ) ) sin ( arctan ( 2 ( x - 2 ) ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (2 (x - 2))) + sec (arctan (2 (x - 2))) ∣) : \frac{1}{4} (2 (x - 2) 4 x^{2} - 16 x + 17 + ln 2 (x - 2) + 4 x^{2} - 16 x + 17)

= \frac{1}{4} (2 (x - 2) 4 x^{2} - 16 x + 17 + ln 2 (x - 2) + 4 x^{2} - 16 x + 17)

Add a constant to the solution

= \frac{1}{4} (2 (x - 2) 4 x^{2} - 16 x + 17 + ln 2 (x - 2) + 4 x^{2} - 16 x + 17) + C

What is the integral of sqrt(1+(2x-4)^2) ?
The integral of sqrt(1+(2x-4)^2) is 1/4 (2(x-2)sqrt(4x^2-16x+17)+ln|2(x-2)+sqrt(4x^2-16x+17)|)+C