What is the integral of 1/((4x^2-4x+27)^{3/2)} ?

The integral of 1/((4x^2-4x+27)^{3/2)} is (2x-1)/(52(4x^2-4x+27)^{1/2)}+C

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integral of 1/((4x^2-4x+27)^{3/2)}

\int \frac{1}{( 4 x ^{2} - 4 x + 27 ) ^{\frac{3}{2}}} d x

\frac{2 x - 1}{52 ( 4 x ^{2} - 4 x + 27 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 4 x ^{2} - 4 x + 27 ) ^{\frac{3}{2}}} d x

Complete the square

4 x^{2} - 4 x + 27 : 4 (x - \frac{1}{2})^{2} + 26

= \int \frac{1}{( 4 ( x - \frac{1}{2} ) ^{2} + 26 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{1}{( 4 u ^{2} + 26 ) ^{\frac{3}{2}}} d u

= \int \frac{1}{( 4 u ^{2} + 26 ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int \frac{1}{52 sec ( v )} d v

= \int \frac{1}{52 sec ( v )} d v

Take the constant out:

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

= \frac{1}{52} \cdot \int \frac{1}{sec ( v )} d v

Rewrite using trig identities

= \frac{1}{52} \cdot \int cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= \frac{1}{52} sin (v)

Substitute back

= \frac{1}{52} sin (arctan (\frac{2}{2 6 ^{\frac{1}{2}}} (x - \frac{1}{2})))

Simplify

\frac{1}{52} sin (arctan (\frac{2}{2 6 ^{\frac{1}{2}}} (x - \frac{1}{2}))) : \frac{2 x - 1}{52 ( 4 x ^{2} - 4 x + 27 ) ^{\frac{1}{2}}}

= \frac{2 x - 1}{52 ( 4 x ^{2} - 4 x + 27 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{2 x - 1}{52 ( 4 x ^{2} - 4 x + 27 ) ^{\frac{1}{2}}} + C

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What is the integral of 1/((4x^2-4x+27)^{3/2)} ?
The integral of 1/((4x^2-4x+27)^{3/2)} is (2x-1)/(52(4x^2-4x+27)^{1/2)}+C