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

The integral of 1/((4x^2-4x-3)^{3/2)} is -(2x-1)/(8(4x^2-4x-3)^{1/2)}+C

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

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

- \frac{2 x - 1}{8 ( 4 x ^{2} - 4 x - 3 ) ^{\frac{1}{2}}} + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{1}{8 ( u ^{2} - 1 ) ^{\frac{3}{2}}} d u

= \int \frac{1}{8 ( u ^{2} - 1 ) ^{\frac{3}{2}}} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{( u ^{2} - 1 ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

: \int cot^{2} (v) sec (v) d v

= \frac{1}{8} \cdot \int cot^{2} (v) sec (v) d v

Simplify

cot^{2} (v) sec (v) : cot (v) csc (v)

= \frac{1}{8} \cdot \int cot (v) csc (v) d v

Apply u-substitution

: \int - 1 d w

= \frac{1}{8} \cdot \int - 1 d w

Integral of a constant:

\int a d x = a x

= \frac{1}{8} (- 1) w

Substitute back

= \frac{1}{8} (- 1) csc (arcsec (x - \frac{1}{2}))

Simplify

\frac{1}{8} (- 1) csc (arcsec (x - \frac{1}{2})) : - \frac{2 x - 1}{8 ( 4 x ^{2} - 4 x - 3 ) ^{\frac{1}{2}}}

= - \frac{2 x - 1}{8 ( 4 x ^{2} - 4 x - 3 ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{2 x - 1}{8 ( 4 x ^{2} - 4 x - 3 ) ^{\frac{1}{2}}} + C

4 + 0.45 \cdot sin (\frac{2 π t}{3.7})

\int 19 x e^{- x^{2}} d x

3 x 3 x 3 x

\frac{d}{d x} (\frac{- 24}{x ^{5}})

\frac{d}{d x} (sin (x^{2}) - sin^{2} (x))

What is the integral of 1/((4x^2-4x-3)^{3/2)} ?
The integral of 1/((4x^2-4x-3)^{3/2)} is -(2x-1)/(8(4x^2-4x-3)^{1/2)}+C