What is the integral of 1/((sqrt(4x^2-9))^3) ?

The integral of 1/((sqrt(4x^2-9))^3) is -x/(9sqrt(4x^2-9))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((sqrt(4x^2-9))^3)

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

- \frac{x}{9 4 x ^{2} - 9} + C

Solution steps

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

Apply exponent rule:

(a^{n})^{m} = a^{n \cdot m}

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

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

Multiply by the conjugate of

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

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

Apply Integral Substitution

= \int \frac{1}{18 ( ( u + 1 ) ( u - 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}{18} \cdot \int \frac{1}{( ( u + 1 ) ( u - 1 ) ) ^{\frac{3}{2}}} d u

Multiply by the conjugate of

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

= \frac{1}{18} \cdot \int \frac{1}{( \frac{u ^{2} - 1}{u - 1} ( u - 1 ) ) ^{\frac{3}{2}}} d u

Multiply

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

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

Simplify

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

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

Apply u-substitution

: \int - 1 d w

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

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

= - \frac{x}{9 4 x ^{2} - 9}

Add a constant to the solution

= - \frac{x}{9 4 x ^{2} - 9} + C

y^{^{'}} = 42 + 20 x + 42 y + 28 x y

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

\int 2 x^{- 2} + \frac{1}{x} d x

\int_{0}^{\frac{1}{2}} 1 arccos (x) d x

e^{z}

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