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

The integral of x(1-4x^2)^{-1/2} is -1/4 (1-4x^2)^{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 x(1-4x^2)^{-1/2}

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{8 u ^{\frac{1}{2}}} d u

= \int - \frac{1}{8 u ^{\frac{1}{2}}} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int \frac{1}{u ^{\frac{1}{2}}} d u

Apply the Power Rule

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

= - \frac{1}{8} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 1 - 4 x^{2}

= - \frac{1}{8} \cdot 2 (1 - 4 x^{2})^{\frac{1}{2}}

Simplify

- \frac{1}{8} \cdot 2 (1 - 4 x^{2})^{\frac{1}{2}} : - \frac{1}{4} (1 - 4 x^{2})^{\frac{1}{2}}

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

Add a constant to the solution

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

\int \frac{32 x ^{2}}{( x - 9 ) ( x + 3 ) ^{2}} d x

f (x) = ln (2 x - 1)

\frac{\partial}{\partial x} (csc (2 x))

\frac{d}{d x} (x - 4 x)

y^{^{'}} (x) = \frac{6 x ^{2}}{6 y ^{2} - 11}, y (1) = 0

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