What is the integral of 4/(1-16x^2) ?

The integral of 4/(1-16x^2) is 1/2 ln|4x+1|-1/2 ln|4x-1|+C

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integral of 4/(1-16x^2)

\int \frac{4}{1 - 16 x ^{2}} d x

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

Solution steps

\int \frac{4}{1 - 16 x ^{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{1}{1 - 16 x ^{2}} d x

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{1 - u ^{2}} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= 4 \cdot \frac{1}{4} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = 4 x

= 4 \cdot \frac{1}{4} (\frac{ln ∣ 4 x + 1 ∣}{2} - \frac{ln ∣ 4 x - 1 ∣}{2})

Simplify

4 \cdot \frac{1}{4} (\frac{ln ∣ 4 x + 1 ∣}{2} - \frac{ln ∣ 4 x - 1 ∣}{2}) : \frac{1}{2} ln ∣ 4 x + 1 ∣ - \frac{1}{2} ln ∣ 4 x - 1 ∣

= \frac{1}{2} ln ∣ 4 x + 1 ∣ - \frac{1}{2} ln ∣ 4 x - 1 ∣

Add a constant to the solution

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

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What is the integral of 4/(1-16x^2) ?
The integral of 4/(1-16x^2) is 1/2 ln|4x+1|-1/2 ln|4x-1|+C