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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int u^{2} + 4 d u

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

Apply Trigonometric Substitution

: \int 4 sec^{3} (v) d v

= \int 4 sec^{3} (v) d v

Take the constant out:

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

= 4 \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= 4 (\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) ∣

= 4 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (\frac{4 x - y}{4 x + y})

(6 + x) y^{^{'}} = 9 y

\int \frac{x + 1}{x ^{2} + 2 x + 4} d x

\frac{d}{d x} (8^{x} - 1 4^{x})

\frac{d y}{d t} = 2 (t + 1) y

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