What is the integral of sqrt(x^2-6x+10) ?

The integral of sqrt(x^2-6x+10) is 1/2 (x-3)sqrt(x^2-6x+10)+1/2 ln|x-3+sqrt(x^2-6x+10)|+C

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integral of sqrt(x^2-6x+10)

\int x^{2} - 6 x + 10 d x

\frac{1}{2} (x - 3) x^{2} - 6 x + 10 + \frac{1}{2} ln x - 3 + x^{2} - 6 x + 10 + C

Solution steps

\int x^{2} - 6 x + 10 d x

Complete the square

x^{2} - 6 x + 10 : (x - 3)^{2} + 1

= \int (x - 3)^{2} + 1 d x

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

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

Apply Integral Reduction

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

Substitute back

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

Simplify

\frac{sec ^{2} ( arctan ( x - 3 ) ) sin ( arctan ( x - 3 ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (x - 3)) + sec (arctan (x - 3)) ∣ : \frac{1}{2} (x - 3) x^{2} - 6 x + 10 + \frac{1}{2} ln x - 3 + x^{2} - 6 x + 10

= \frac{1}{2} (x - 3) x^{2} - 6 x + 10 + \frac{1}{2} ln x - 3 + x^{2} - 6 x + 10

Add a constant to the solution

= \frac{1}{2} (x - 3) x^{2} - 6 x + 10 + \frac{1}{2} ln x - 3 + x^{2} - 6 x + 10 + C

y^{^{''}} - 5 y^{^{'}} = (8 t - 7) sin (2 t)

x \to 7 lim (x (x - 2))

\frac{d}{d x} (cos (5 x^{2} + 7 x - 1))

2 x \cdot cos (x^{2})

\int \frac{1}{1 0 ^{x}} d x

What is the integral of sqrt(x^2-6x+10) ?
The integral of sqrt(x^2-6x+10) is 1/2 (x-3)sqrt(x^2-6x+10)+1/2 ln|x-3+sqrt(x^2-6x+10)|+C