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

The integral of x/(1-4x^2) is -1/8 ln|1-4x^2|+C

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

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

- \frac{1}{8} ln 1 - 4 x^{2} + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{8} ln ∣ u ∣

Substitute back

u = 1 - 4 x^{2}

= - \frac{1}{8} ln 1 - 4 x^{2}

Add a constant to the solution

= - \frac{1}{8} ln 1 - 4 x^{2} + C

2^{x} lo g_{2} (x)

\int \frac{3 x - 1}{x + 1} d x

\frac{\partial}{\partial x} (y - x^{2} - x y^{2})

\int_{1}^{3} 3 x^{2} + \frac{1}{12 x ^{2}} d x

x \to 3 lim (- 4)

What is the integral of x/(1-4x^2) ?
The integral of x/(1-4x^2) is -1/8 ln|1-4x^2|+C