What is the integral of 1/((4x^2-24x+27)^{3/2)} ?

The integral of 1/((4x^2-24x+27)^{3/2)} is -(x-3)/(9(4x^2-24x+27)^{1/2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((4x^2-24x+27)^{3/2)}

\int \frac{1}{( 4 x ^{2} - 24 x + 27 ) ^{\frac{3}{2}}} d x

- \frac{x - 3}{9 ( 4 x ^{2} - 24 x + 27 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 4 x ^{2} - 24 x + 27 ) ^{\frac{3}{2}}} d x

Complete the square

4 x^{2} - 24 x + 27 : 4 (x - 3)^{2} - 9

= \int \frac{1}{( 4 ( x - 3 ) ^{2} - 9 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{1}{( 4 u ^{2} - 9 ) ^{\frac{3}{2}}} d u

= \int \frac{1}{( 4 u ^{2} - 9 ) ^{\frac{3}{2}}} d u

Multiply by the conjugate of

4 u^{2} - 9 : \frac{16 u ^{4} - 81}{4 u ^{2} + 9}

= \int \frac{1}{( \frac{16 u ^{4} - 81}{4 u ^{2} + 9} ) ^{\frac{3}{2}}} d u

Apply Integral Substitution

= \int \frac{1}{18 ( ( 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}{18} \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}{18} \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}{18} \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}{18} \cdot \int cot^{2} (w) sec (w) d w

Simplify

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

= \frac{1}{18} \cdot \int cot (w) csc (w) d w

Apply u-substitution

: \int - 1 d t

= \frac{1}{18} \cdot \int - 1 d t

Integral of a constant:

\int a d x = a x

= \frac{1}{18} (- 1) t

Substitute back

= \frac{1}{18} (- 1) csc (arcsec (\frac{2}{3} (x - 3)))

Simplify

\frac{1}{18} (- 1) csc (arcsec (\frac{2}{3} (x - 3))) : - \frac{x - 3}{9 ( 4 x ^{2} - 24 x + 27 ) ^{\frac{1}{2}}}

= - \frac{x - 3}{9 ( 4 x ^{2} - 24 x + 27 ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{x - 3}{9 ( 4 x ^{2} - 24 x + 27 ) ^{\frac{1}{2}}} + C

\int \frac{1}{5 x ( 1 + 5 x )} d x

\frac{d y}{d x} = (1 - \frac{2}{x ^{2}}) (y - 1)

\frac{\partial}{\partial y} (x^{3} y + x cos (y))

\int_{- 4}^{4} 3 (16 - y^{2}) d y

\frac{d}{d x} (\frac{2 x ^{2} + x - 3}{2 x + 7})

What is the integral of 1/((4x^2-24x+27)^{3/2)} ?
The integral of 1/((4x^2-24x+27)^{3/2)} is -(x-3)/(9(4x^2-24x+27)^{1/2)}+C