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

The integral of 1/((1-2x)^4sqrt(4x^2-4x-4)) is -((8x^2-8x+7)sqrt(x^2-x-1))/(75(-2x+1)^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{( 8 x ^{2} - 8 x + 7 ) x ^{2} - x - 1}{75 ( - 2 x + 1 ) ^{3}} + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{2 u ^{4} u ^{2} - 5} d u

= \int - \frac{1}{2 u ^{4} u ^{2} - 5} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{1}{u ^{4} u ^{2} - 5} d u

Apply Trigonometric Substitution

: \int \frac{1}{25 sec ^{3} ( v )} d v

= - \frac{1}{2} \cdot \int \frac{1}{25 sec ^{3} ( v )} d v

Take the constant out:

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

= - \frac{1}{2} \cdot \frac{1}{25} \cdot \int \frac{1}{sec ^{3} ( v )} d v

Rewrite using trig identities

= - \frac{1}{2} \cdot \frac{1}{25} \cdot \int (1 - sin^{2} (v)) cos (v) d v

Apply u-substitution

: \int 1 - w^{2} d w

= - \frac{1}{2} \cdot \frac{1}{25} \cdot \int 1 - w^{2} d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \frac{1}{2} \cdot \frac{1}{25} (\int 1 d w - \int w^{2} d w)

\int 1 d w = w

\int w^{2} d w = \frac{w ^{3}}{3}

= - \frac{1}{2} \cdot \frac{1}{25} (w - \frac{w ^{3}}{3})

Substitute back

= - \frac{1}{2} \cdot \frac{1}{25} sin (arcsec (\frac{1}{5} (1 - 2 x))) - \frac{sin ^{3} ( arcsec ( \frac{1}{5} ( 1 - 2 x ) ) )}{3}

Simplify

- \frac{1}{2} \cdot \frac{1}{25} sin (arcsec (\frac{1}{5} (1 - 2 x))) - \frac{sin ^{3} ( arcsec ( \frac{1}{5} ( 1 - 2 x ) ) )}{3} : - \frac{( 8 x ^{2} - 8 x + 7 ) x ^{2} - x - 1}{75 ( - 2 x + 1 ) ^{3}}

= - \frac{( 8 x ^{2} - 8 x + 7 ) x ^{2} - x - 1}{75 ( - 2 x + 1 ) ^{3}}

Add a constant to the solution

= - \frac{( 8 x ^{2} - 8 x + 7 ) x ^{2} - x - 1}{75 ( - 2 x + 1 ) ^{3}} + C

\frac{d}{d x} (\frac{3}{1 + 3 x})

\int \frac{4}{65} x^{3} d x

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

f (x) = 2 x, (0, 3)

e^{x} - 11

What is the integral of 1/((1-2x)^4sqrt(4x^2-4x-4)) ?
The integral of 1/((1-2x)^4sqrt(4x^2-4x-4)) is -((8x^2-8x+7)sqrt(x^2-x-1))/(75(-2x+1)^3)+C