What is the integral of 1/((4x^2-24x+7)^{3/2)} ?

The integral of 1/((4x^2-24x+7)^{3/2)} is -(x-3)/(29(4x^2-24x+7)^{1/2)}+C

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integral of 1/((4x^2-24x+7)^{3/2)}

\int \frac{1}{( 4 x ^{2} - 24 x + 7 ) ^{\frac{3}{2}}} d x

- \frac{x - 3}{29 ( 4 x ^{2} - 24 x + 7 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 4 x ^{2} - 24 x + 7 ) ^{\frac{3}{2}}} d x

Complete the square

4 x^{2} - 24 x + 7 : 4 (x - 3)^{2} - 29

= \int \frac{1}{( 4 ( x - 3 ) ^{2} - 29 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{1}{( 4 u ^{2} - 29 ) ^{\frac{3}{2}}} d u

= \int \frac{1}{( 4 u ^{2} - 29 ) ^{\frac{3}{2}}} d u

Multiply by the conjugate of

4 u^{2} - 29 : \frac{16 u ^{4} - 841}{4 u ^{2} + 29}

= \int \frac{1}{( \frac{16 u ^{4} - 841}{4 u ^{2} + 29} ) ^{\frac{3}{2}}} d u

Apply Integral Substitution

= \int \frac{1}{58 ( ( v + 1 ) ( v - 1 ) ) ^{\frac{3}{2}}} d v

Take the constant out:

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

= \frac{1}{58} \cdot \int \frac{1}{( ( v + 1 ) ( v - 1 ) ) ^{\frac{3}{2}}} d v

Multiply by the conjugate of

v + 1 : \frac{v ^{2} - 1}{v - 1}

= \frac{1}{58} \cdot \int \frac{1}{( \frac{v ^{2} - 1}{v - 1} ( v - 1 ) ) ^{\frac{3}{2}}} d v

Multiply

\frac{v ^{2} - 1}{v - 1} (v - 1) : v^{2} - 1

= \frac{1}{58} \cdot \int \frac{1}{( v ^{2} - 1 ) ^{\frac{3}{2}}} d v

Apply Trigonometric Substitution

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

= \frac{1}{58} \cdot \int cot^{2} (w) sec (w) d w

Simplify

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

= \frac{1}{58} \cdot \int cot (w) csc (w) d w

Apply u-substitution

: \int - 1 d t

= \frac{1}{58} \cdot \int - 1 d t

Integral of a constant:

\int a d x = a x

= \frac{1}{58} (- 1) t

Substitute back

= \frac{1}{58} (- 1) csc (arcsec (\frac{2}{2 9 ^{\frac{1}{2}}} (x - 3)))

Simplify

\frac{1}{58} (- 1) csc (arcsec (\frac{2}{2 9 ^{\frac{1}{2}}} (x - 3))) : - \frac{x - 3}{29 ( 4 x ^{2} - 24 x + 7 ) ^{\frac{1}{2}}}

= - \frac{x - 3}{29 ( 4 x ^{2} - 24 x + 7 ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{x - 3}{29 ( 4 x ^{2} - 24 x + 7 ) ^{\frac{1}{2}}} + C

\int 2 sin (x) - sec^{2} (x) d x

\frac{\partial}{\partial x} (6 x y - 12)

\int 2 tan (x) sec^{3} (x) d x

\frac{d}{d x} (\frac{π}{( 2 x ) ^{5}})

y = \frac{x ^{2} - 4 x + 3}{x ^{2}}

What is the integral of 1/((4x^2-24x+7)^{3/2)} ?
The integral of 1/((4x^2-24x+7)^{3/2)} is -(x-3)/(29(4x^2-24x+7)^{1/2)}+C