What is the integral of sqrt(x^2-4x+29) ?

The integral of sqrt(x^2-4x+29) is 25(1/50 (x-2)sqrt(x^2-4x+29)+1/2 ln((|x-2+sqrt(x^2-4x+29)|)/5))+C

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integral of sqrt(x^2-4x+29)

\int x^{2} - 4 x + 29 d x

25 (\frac{1}{50} (x - 2) x^{2} - 4 x + 29 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 29}{5})) + C

Solution steps

\int x^{2} - 4 x + 29 d x

Complete the square

x^{2} - 4 x + 29 : (x - 2)^{2} + 25

= \int (x - 2)^{2} + 25 d x

Apply u-substitution

: \int u^{2} + 25 d u

= \int u^{2} + 25 d u

Apply Trigonometric Substitution

: \int 25 sec^{3} (v) d v

= \int 25 sec^{3} (v) d v

Take the constant out:

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

= 25 \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= 25 (\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) ∣

= 25 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= 25 (\frac{sec ^{2} ( arctan ( \frac{1}{5} ( x - 2 ) ) ) sin ( arctan ( \frac{1}{5} ( x - 2 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{5} (x - 2))) + sec (arctan (\frac{1}{5} (x - 2))))

Simplify

25 (\frac{sec ^{2} ( arctan ( \frac{1}{5} ( x - 2 ) ) ) sin ( arctan ( \frac{1}{5} ( x - 2 ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{5} (x - 2))) + sec (arctan (\frac{1}{5} (x - 2)))) : 25 (\frac{1}{50} (x - 2) x^{2} - 4 x + 29 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 29}{5}))

= 25 (\frac{1}{50} (x - 2) x^{2} - 4 x + 29 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 29}{5}))

Add a constant to the solution

= 25 (\frac{1}{50} (x - 2) x^{2} - 4 x + 29 + \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 29}{5})) + C

\frac{d}{d x} ((1 + x)^{2} ln (1 + x))

y^{10} y^{^{'}} = 13 cos (π x)

\int 7 - 3 (1 + x^{2})^{- 1} d x

6 x^{^{''}} + 1960 x = 0, x (0) = - 3, x^{^{'}} (0) = 0

f (x) = (3)^{x}, at x = 8

What is the integral of sqrt(x^2-4x+29) ?
The integral of sqrt(x^2-4x+29) is 25(1/50 (x-2)sqrt(x^2-4x+29)+1/2 ln((|x-2+sqrt(x^2-4x+29)|)/5))+C