What is the integral of sqrt(4x^2-4x+10) ?

The integral of sqrt(4x^2-4x+10) is 1/4 ((2x-1)sqrt(4x^2-4x+10)+9ln((|2x+sqrt(4x^2-4x+10)-1|)/3))+C

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integral of sqrt(4x^2-4x+10)

\int 4 x^{2} - 4 x + 10 d x

\frac{1}{4} ((2 x - 1) 4 x^{2} - 4 x + 10 + 9 ln (\frac{2 x + 4 x ^{2} - 4 x + 10 - 1}{3})) + C

Solution steps

\int 4 x^{2} - 4 x + 10 d x

Complete the square

4 x^{2} - 4 x + 10 : 4 (x - \frac{1}{2})^{2} + 9

= \int 4 (x - \frac{1}{2})^{2} + 9 d x

Apply u-substitution

: \int 4 u^{2} + 9 d u

= \int 4 u^{2} + 9 d u

Apply Trigonometric Substitution

: \int \frac{9}{2} sec^{3} (v) d v

= \int \frac{9}{2} sec^{3} (v) d v

Take the constant out:

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

= \frac{9}{2} \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= \frac{9}{2} (\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{9}{2} (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= \frac{9}{2} (\frac{sec ^{2} ( arctan ( \frac{2}{3} ( x - \frac{1}{2} ) ) ) sin ( arctan ( \frac{2}{3} ( x - \frac{1}{2} ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{3} (x - \frac{1}{2}))) + sec (arctan (\frac{2}{3} (x - \frac{1}{2}))))

Simplify

\frac{9}{2} (\frac{sec ^{2} ( arctan ( \frac{2}{3} ( x - \frac{1}{2} ) ) ) sin ( arctan ( \frac{2}{3} ( x - \frac{1}{2} ) ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{3} (x - \frac{1}{2}))) + sec (arctan (\frac{2}{3} (x - \frac{1}{2})))) : \frac{1}{4} ((2 x - 1) 4 x^{2} - 4 x + 10 + 9 ln (\frac{2 x + 4 x ^{2} - 4 x + 10 - 1}{3}))

= \frac{1}{4} ((2 x - 1) 4 x^{2} - 4 x + 10 + 9 ln (\frac{2 x + 4 x ^{2} - 4 x + 10 - 1}{3}))

Add a constant to the solution

= \frac{1}{4} ((2 x - 1) 4 x^{2} - 4 x + 10 + 9 ln (\frac{2 x + 4 x ^{2} - 4 x + 10 - 1}{3})) + C

\int_{0}^{2} f (x^{3}) d x

\frac{d ^{2} x}{d x ^{2}}

x y^{^{'}} + 2 y = \frac{3 e ^{- 4 x}}{x}

x \to - 1 + lim (\frac{x}{x ^{2} - 1})

\int_{0}^{4} 5 e^{- x^{2}} d x

What is the integral of sqrt(4x^2-4x+10) ?
The integral of sqrt(4x^2-4x+10) is 1/4 ((2x-1)sqrt(4x^2-4x+10)+9ln((|2x+sqrt(4x^2-4x+10)-1|)/3))+C