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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{1}{1 - 2 x ^{2}} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{u}

= - 4 (- \frac{1}{4} (- \frac{1}{u}))

Substitute back

u = 1 - 2 x^{2}

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

Simplify

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

= - \frac{1}{1 - 2 x ^{2}}

Add a constant to the solution

= - \frac{1}{1 - 2 x ^{2}} + C

ln (\frac{1 + x}{1 - x}), 0

y^{^{'}} + x y^{2} = 0

x \to 0 lim (x + a)

x \to 3.1 lim (\frac{5 x}{6 - 2 x})

\frac{\partial}{\partial y} (3 x^{2} y^{4} + 3)

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