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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

Take the constant out:

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

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

Apply u-substitution

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{u u ^{2} - 4} d u

Apply Trigonometric Substitution

= 3 \cdot \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

= 3 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 3 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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