What is the integral of sqrt((x^2)/4-4x+17) ?

The integral of sqrt((x^2)/4-4x+17) is ln(1/2 |x-8+sqrt(x^2-16x+68)|)+1/4 (x-8)sqrt(x^2-16x+68)+C

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integral of sqrt((x^2)/4-4x+17)

\int \frac{x ^{2}}{4} - 4 x + 17 d x

ln (\frac{1}{2} x - 8 + x^{2} - 16 x + 68) + \frac{1}{4} (x - 8) x^{2} - 16 x + 68 + C

Solution steps

\int \frac{x ^{2}}{4} - 4 x + 17 d x

Simplify

\frac{x ^{2}}{4} - 4 x + 17 : \frac{1}{2} x^{2} - 16 x + 68

= \int \frac{1}{2} x^{2} - 16 x + 68 d x

Take the constant out:

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

= \frac{1}{2} \cdot \int x^{2} - 16 x + 68 d x

Complete the square

x^{2} - 16 x + 68 : (x - 8)^{2} + 4

= \frac{1}{2} \cdot \int (x - 8)^{2} + 4 d x

Apply u-substitution

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

= \frac{1}{2} \cdot \int u^{2} + 4 d u

Apply Trigonometric Substitution

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

= \frac{1}{2} \cdot \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

= \frac{1}{2} \cdot 4 \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= \frac{1}{2} \cdot 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) ∣

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

Substitute back

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

Simplify

\frac{1}{2} \cdot 4 (\frac{sec ^{2} ( arctan ( \frac{1}{2} ( x - 8 ) ) ) sin ( arctan ( \frac{1}{2} ( x - 8 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{2} (x - 8))) + sec (arctan (\frac{1}{2} (x - 8)))) : ln (\frac{1}{2} x - 8 + x^{2} - 16 x + 68) + \frac{1}{4} (x - 8) x^{2} - 16 x + 68

= ln (\frac{1}{2} x - 8 + x^{2} - 16 x + 68) + \frac{1}{4} (x - 8) x^{2} - 16 x + 68

Add a constant to the solution

= ln (\frac{1}{2} x - 8 + x^{2} - 16 x + 68) + \frac{1}{4} (x - 8) x^{2} - 16 x + 68 + C

\frac{d}{d x} (in (x + 3))

\frac{( x ^{2} + 2 y ^{2} ) d x}{d y} = x y

\int_{5}^{8} \frac{y}{y ^{2} - y - 2} d y

e^{x}, e^{5 x}, 0, 1

What is the integral of sqrt((x^2)/4-4x+17) ?
The integral of sqrt((x^2)/4-4x+17) is ln(1/2 |x-8+sqrt(x^2-16x+68)|)+1/4 (x-8)sqrt(x^2-16x+68)+C