What is the integral of 1/(sqrt(4x^2-16x+52)) ?

The integral of 1/(sqrt(4x^2-16x+52)) is 1/2 ln((|x-2+sqrt(x^2-4x+13)|)/3)+C

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integral of 1/(sqrt(4x^2-16x+52))

\int \frac{1}{4 x ^{2} - 16 x + 52} d x

\frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 13}{3}) + C

Solution steps

\int \frac{1}{4 x ^{2} - 16 x + 52} d x

Complete the square

4 x^{2} - 16 x + 52 : 4 (x - 2)^{2} + 36

= \int \frac{1}{4 ( x - 2 ) ^{2} + 36} d x

Apply u-substitution

: \int \frac{1}{2 u ^{2} + 9} d u

= \int \frac{1}{2 u ^{2} + 9} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int sec (v) d v

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

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

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

Substitute back

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

Simplify

\frac{1}{2} ln tan (arctan (\frac{1}{3} (x - 2))) + sec (arctan (\frac{1}{3} (x - 2))) : \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 13}{3})

= \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 13}{3})

Add a constant to the solution

= \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 13}{3}) + C

\frac{\partial ^{2}}{\partial y ^{2}} (y^{4} cos (2 x))

y^{^{'}} = 10 y

\frac{d}{d x} (x - \frac{1}{7} x)

\frac{\partial}{\partial x} ((e^{- x}) sin (yz))

\frac{d}{d x} (y cos (2 x))

What is the integral of 1/(sqrt(4x^2-16x+52)) ?
The integral of 1/(sqrt(4x^2-16x+52)) is 1/2 ln((|x-2+sqrt(x^2-4x+13)|)/3)+C