What is the integral of (4x^2+48)/(16-x^4) ?

The integral of (4x^2+48)/(16-x^4) is 2arctan(x/2)+2ln|x+2|-2ln|x-2|+C

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integral of (4x^2+48)/(16-x^4)

\int \frac{4 x ^{2} + 48}{16 - x ^{4}} d x

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

Solution steps

\int \frac{4 x ^{2} + 48}{16 - x ^{4}} d x

Take the partial fraction of

\frac{4 x ^{2} + 48}{16 - x ^{4}} : \frac{4}{x ^{2} + 4} + \frac{2}{x + 2} - \frac{2}{x - 2}

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

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

\int \frac{2}{x + 2} d x = 2 ln ∣ x + 2 ∣

\int \frac{2}{x - 2} d x = 2 ln ∣ x - 2 ∣

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

Add a constant to the solution

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

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

\int x^{2} + 6 x d x

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

\int e^{x} - 3 x^{2} + sin (x) d x

\int e^{t} - 3 d t

What is the integral of (4x^2+48)/(16-x^4) ?
The integral of (4x^2+48)/(16-x^4) is 2arctan(x/2)+2ln|x+2|-2ln|x-2|+C