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

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

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

Complete the square

x^{2} - 2 x + 5 : (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

\int \frac{1}{x ^{2} + 3} d x

\frac{1}{x - 8}

r = cos (2 θ)

\frac{\partial}{\partial y} (y^{4} sin (3 x))

\int \frac{1}{x ^{2} + 2 x + 6} d x

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