What is the integral of sqrt(x^2-6x+34) ?

The integral of sqrt(x^2-6x+34) is 25(1/50 (x-3)sqrt(x^2-6x+34)+1/2 ln((|x-3+sqrt(x^2-6x+34)|)/5))+C

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integral of sqrt(x^2-6x+34)

\int x^{2} - 6 x + 34 d x

25 (\frac{1}{50} (x - 3) x^{2} - 6 x + 34 + \frac{1}{2} ln (\frac{x - 3 + x ^{2} - 6 x + 34}{5})) + C

Solution steps

\int x^{2} - 6 x + 34 d x

Complete the square

x^{2} - 6 x + 34 : (x - 3)^{2} + 25

= \int (x - 3)^{2} + 25 d x

Apply u-substitution

: \int u^{2} + 25 d u

= \int u^{2} + 25 d u

Apply Trigonometric Substitution

: \int 25 sec^{3} (v) d v

= \int 25 sec^{3} (v) d v

Take the constant out:

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

= 25 \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= 25 (\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) ∣

= 25 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 25 (\frac{sec ^{2} ( arctan ( \frac{1}{5} ( x - 3 ) ) ) sin ( arctan ( \frac{1}{5} ( x - 3 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{5} (x - 3))) + sec (arctan (\frac{1}{5} (x - 3))))

Simplify

25 (\frac{sec ^{2} ( arctan ( \frac{1}{5} ( x - 3 ) ) ) sin ( arctan ( \frac{1}{5} ( x - 3 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{5} (x - 3))) + sec (arctan (\frac{1}{5} (x - 3)))) : 25 (\frac{1}{50} (x - 3) x^{2} - 6 x + 34 + \frac{1}{2} ln (\frac{x - 3 + x ^{2} - 6 x + 34}{5}))

= 25 (\frac{1}{50} (x - 3) x^{2} - 6 x + 34 + \frac{1}{2} ln (\frac{x - 3 + x ^{2} - 6 x + 34}{5}))

Add a constant to the solution

= 25 (\frac{1}{50} (x - 3) x^{2} - 6 x + 34 + \frac{1}{2} ln (\frac{x - 3 + x ^{2} - 6 x + 34}{5})) + C

x \to 2 + lim (in t x)

\frac{d}{d θ} (sin^{2} (θ))

y^{^{''}} + 49 y = 3 tan (7 x)

f (x) = 2 x + 7, at x = 9

y^{^{'}} + 0.7 y - 18.2 = 0

What is the integral of sqrt(x^2-6x+34) ?
The integral of sqrt(x^2-6x+34) is 25(1/50 (x-3)sqrt(x^2-6x+34)+1/2 ln((|x-3+sqrt(x^2-6x+34)|)/5))+C