What is the integral of sqrt(x^2-4x+40) ?

The integral of sqrt(x^2-4x+40) is 36(1/72 (x-2)sqrt(x^2-4x+40)+1/2 ln((|x-2+sqrt(x^2-4x+40)|)/6))+C

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integral of sqrt(x^2-4x+40)

\int x^{2} - 4 x + 40 d x

36 (\frac{1}{72} (x - 2) x^{2} - 4 x + 40 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 40}{6})) + C

Solution steps

\int x^{2} - 4 x + 40 d x

Complete the square

x^{2} - 4 x + 40 : (x - 2)^{2} + 36

= \int (x - 2)^{2} + 36 d x

Apply u-substitution

: \int u^{2} + 36 d u

= \int u^{2} + 36 d u

Apply Trigonometric Substitution

: \int 36 sec^{3} (v) d v

= \int 36 sec^{3} (v) d v

Take the constant out:

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

= 36 \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= 36 (\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) ∣

= 36 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 36 (\frac{sec ^{2} ( arctan ( \frac{1}{6} ( x - 2 ) ) ) sin ( arctan ( \frac{1}{6} ( x - 2 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{6} (x - 2))) + sec (arctan (\frac{1}{6} (x - 2))))

Simplify

36 (\frac{sec ^{2} ( arctan ( \frac{1}{6} ( x - 2 ) ) ) sin ( arctan ( \frac{1}{6} ( x - 2 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{6} (x - 2))) + sec (arctan (\frac{1}{6} (x - 2)))) : 36 (\frac{1}{72} (x - 2) x^{2} - 4 x + 40 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 40}{6}))

= 36 (\frac{1}{72} (x - 2) x^{2} - 4 x + 40 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 40}{6}))

Add a constant to the solution

= 36 (\frac{1}{72} (x - 2) x^{2} - 4 x + 40 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 40}{6})) + C

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = - 2, y^{^{'}} (0) = 3

f (x) = (12 x^{3} - 9 x)^{\frac{4}{3}}

\frac{d}{d x} (x^{3} - 9 x)

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

f (x) = \frac{4}{x}, at x = - 2

What is the integral of sqrt(x^2-4x+40) ?
The integral of sqrt(x^2-4x+40) is 36(1/72 (x-2)sqrt(x^2-4x+40)+1/2 ln((|x-2+sqrt(x^2-4x+40)|)/6))+C