What is the integral of 1/(sqrt(4x^2-12x+7)) ?

The integral of 1/(sqrt(4x^2-12x+7)) is 1/2 ln|sqrt(2(x^2-3x+9/4)-1)+sqrt(2)(x-3/2)|+C

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integral of 1/(sqrt(4x^2-12x+7))

\int \frac{1}{4 x ^{2} - 12 x + 7} d x

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

Solution steps

\int \frac{1}{4 x ^{2} - 12 x + 7} d x

Complete the square

4 x^{2} - 12 x + 7 : 4 (x - \frac{3}{2})^{2} - 2

= \int \frac{1}{4 ( x - \frac{3}{2} ) ^{2} - 2} d x

Apply u-substitution

: \int \frac{1}{4 u ^{2} - 2} d u

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

= \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 (arcsec (\frac{2}{2} (x - \frac{3}{2}))) + sec (arcsec (\frac{2}{2} (x - \frac{3}{2})))

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 1/(sqrt(4x^2-12x+7)) ?
The integral of 1/(sqrt(4x^2-12x+7)) is 1/2 ln|sqrt(2(x^2-3x+9/4)-1)+sqrt(2)(x-3/2)|+C