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

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

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

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

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

Solution steps

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

\frac{x ^{4}}{x ^{4} - 16} = \frac{16}{x ^{4} - 16} + 1

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

\int \frac{16}{x ^{4} - 16} d x = - arctan (\frac{x}{2}) - \frac{1}{2} ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x - 2 ∣

\int 1 d x = x

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

Add a constant to the solution

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

y^{^{''}} + 9 y = 2 cos (3 x) + 3 sin (3 x)

f (x) = 4 sin (x), at x = \frac{π}{2}

y y^{^{'}} = y^{2} + 1

f (x) = csc^{3} (5 x^{2} + 1)

\int cot^{4} (\frac{1}{2}) t d t

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