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

The integral of 1/(16x^4-1) is -1/4 arctan(2x)-1/8 ln|2x+1|+1/8 ln|2x-1|+C

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

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

- \frac{1}{4} arctan (2 x) - \frac{1}{8} ln ∣ 2 x + 1 ∣ + \frac{1}{8} ln ∣ 2 x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{16 x ^{4} - 1} : - \frac{1}{2 ( 4 x ^{2} + 1 )} - \frac{1}{4 ( 2 x + 1 )} + \frac{1}{4 ( 2 x - 1 )}

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

Apply the Sum Rule:

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

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

\int \frac{1}{2 ( 4 x ^{2} + 1 )} d x = \frac{1}{4} arctan (2 x)

\int \frac{1}{4 ( 2 x + 1 )} d x = \frac{1}{8} ln ∣ 2 x + 1 ∣

\int \frac{1}{4 ( 2 x - 1 )} d x = \frac{1}{8} ln ∣ 2 x - 1 ∣

= - \frac{1}{4} arctan (2 x) - \frac{1}{8} ln ∣ 2 x + 1 ∣ + \frac{1}{8} ln ∣ 2 x - 1 ∣

Add a constant to the solution

= - \frac{1}{4} arctan (2 x) - \frac{1}{8} ln ∣ 2 x + 1 ∣ + \frac{1}{8} ln ∣ 2 x - 1 ∣ + C

\frac{d}{d x} (e^{r (t, x) t})

\int \frac{3 e ^{3 t}}{3 + e ^{3 t}} d t

y = ((25 - x^{2})^{\frac{1}{2}}), y = \frac{4}{3} x, x = 5

(\frac{1 + x ^{2}}{1 - x ^{2}})^{17}

\frac{d y}{d x} = sin (17 x)

What is the integral of 1/(16x^4-1) ?
The integral of 1/(16x^4-1) is -1/4 arctan(2x)-1/8 ln|2x+1|+1/8 ln|2x-1|+C