What is the integral of 1/((x-3)sqrt(4x^2-24x+20)) ?

The integral of 1/((x-3)sqrt(4x^2-24x+20)) is 1/4 arcsec(1/2 (x-3))+C

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integral of 1/((x-3)sqrt(4x^2-24x+20))

\int \frac{1}{( x - 3 ) 4 x ^{2} - 24 x + 20} d x

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

Solution steps

\int \frac{1}{( x - 3 ) 4 x ^{2} - 24 x + 20} d x

4 x^{2} - 24 x + 20 = 2 x^{2} - 6 x + 5

= \int \frac{1}{2 ( x - 3 ) x ^{2} - 6 x + 5} d x

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}{( x - 3 ) x ^{2} - 6 x + 5} d x

Apply u-substitution

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

= \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} (x - 3))

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 1/((x-3)sqrt(4x^2-24x+20)) ?
The integral of 1/((x-3)sqrt(4x^2-24x+20)) is 1/4 arcsec(1/2 (x-3))+C