What is the integral of sqrt(x^2-4x+13) ?

The integral of sqrt(x^2-4x+13) is 9(1/18 (x-2)sqrt(x^2-4x+13)+1/2 ln((|x-2+sqrt(x^2-4x+13)|)/3))+C

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integral of sqrt(x^2-4x+13)

\int x^{2} - 4 x + 13 d x

9 (\frac{1}{18} (x - 2) x^{2} - 4 x + 13 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 13}{3})) + C

Solution steps

\int x^{2} - 4 x + 13 d x

Complete the square

x^{2} - 4 x + 13 : (x - 2)^{2} + 9

= \int (x - 2)^{2} + 9 d x

Apply u-substitution

: \int u^{2} + 9 d u

= \int u^{2} + 9 d u

Apply Trigonometric Substitution

: \int 9 sec^{3} (v) d v

= \int 9 sec^{3} (v) d v

Take the constant out:

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

= 9 \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= 9 (\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) ∣

= 9 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 9 (\frac{sec ^{2} ( arctan ( \frac{1}{3} ( x - 2 ) ) ) sin ( arctan ( \frac{1}{3} ( x - 2 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{3} (x - 2))) + sec (arctan (\frac{1}{3} (x - 2))))

Simplify

9 (\frac{sec ^{2} ( arctan ( \frac{1}{3} ( x - 2 ) ) ) sin ( arctan ( \frac{1}{3} ( x - 2 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{3} (x - 2))) + sec (arctan (\frac{1}{3} (x - 2)))) : 9 (\frac{1}{18} (x - 2) x^{2} - 4 x + 13 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 13}{3}))

= 9 (\frac{1}{18} (x - 2) x^{2} - 4 x + 13 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 13}{3}))

Add a constant to the solution

= 9 (\frac{1}{18} (x - 2) x^{2} - 4 x + 13 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 13}{3})) + C

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What is the integral of sqrt(x^2-4x+13) ?
The integral of sqrt(x^2-4x+13) is 9(1/18 (x-2)sqrt(x^2-4x+13)+1/2 ln((|x-2+sqrt(x^2-4x+13)|)/3))+C