What is the integral of 1/((2x-3)sqrt(5-12x+4x^2)) ?

The integral of 1/((2x-3)sqrt(5-12x+4x^2)) is 1/4 arcsec(x-3/2)+C

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integral of 1/((2x-3)sqrt(5-12x+4x^2))

\int \frac{1}{( 2 x - 3 ) 5 - 12 x + 4 x ^{2}} d x

\frac{1}{4} arcsec (x - \frac{3}{2}) + C

Solution steps

\int \frac{1}{( 2 x - 3 ) 5 - 12 x + 4 x ^{2}} d x

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= \frac{1}{2} \cdot \int \frac{1}{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 \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} \cdot 1 \cdot arcsec (\frac{1}{2} (2 x - 3))

Simplify

\frac{1}{2} \cdot \frac{1}{2} \cdot 1 \cdot arcsec (\frac{1}{2} (2 x - 3)) : \frac{1}{4} arcsec (x - \frac{3}{2})

= \frac{1}{4} arcsec (x - \frac{3}{2})

Add a constant to the solution

= \frac{1}{4} arcsec (x - \frac{3}{2}) + C

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What is the integral of 1/((2x-3)sqrt(5-12x+4x^2)) ?
The integral of 1/((2x-3)sqrt(5-12x+4x^2)) is 1/4 arcsec(x-3/2)+C