What is the integral of (64x^2)/(x^4-40x^2+144) ?

The integral of (64x^2)/(x^4-40x^2+144) is 64(1/32 ln|x+2|-1/32 ln|x-2|-3/32 ln|x+6|+3/32 ln|x-6|)+C

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integral of (64x^2)/(x^4-40x^2+144)

\int \frac{64 x ^{2}}{x ^{4} - 40 x ^{2} + 144} d x

64 (\frac{1}{32} ln ∣ x + 2 ∣ - \frac{1}{32} ln ∣ x - 2 ∣ - \frac{3}{32} ln ∣ x + 6 ∣ + \frac{3}{32} ln ∣ x - 6 ∣) + C

Solution steps

\int \frac{64 x ^{2}}{x ^{4} - 40 x ^{2} + 144} d x

Take the constant out:

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

= 64 \cdot \int \frac{x ^{2}}{x ^{4} - 40 x ^{2} + 144} d x

Take the partial fraction of

\frac{x ^{2}}{x ^{4} - 40 x ^{2} + 144} : \frac{1}{32 ( x + 2 )} - \frac{1}{32 ( x - 2 )} - \frac{3}{32 ( x + 6 )} + \frac{3}{32 ( x - 6 )}

= 64 \cdot \int \frac{1}{32 ( x + 2 )} - \frac{1}{32 ( x - 2 )} - \frac{3}{32 ( x + 6 )} + \frac{3}{32 ( x - 6 )} 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

= 64 (\int \frac{1}{32 ( x + 2 )} d x - \int \frac{1}{32 ( x - 2 )} d x - \int \frac{3}{32 ( x + 6 )} d x + \int \frac{3}{32 ( x - 6 )} d x)

\int \frac{1}{32 ( x + 2 )} d x = \frac{1}{32} ln ∣ x + 2 ∣

\int \frac{1}{32 ( x - 2 )} d x = \frac{1}{32} ln ∣ x - 2 ∣

\int \frac{3}{32 ( x + 6 )} d x = \frac{3}{32} ln ∣ x + 6 ∣

\int \frac{3}{32 ( x - 6 )} d x = \frac{3}{32} ln ∣ x - 6 ∣

= 64 (\frac{1}{32} ln ∣ x + 2 ∣ - \frac{1}{32} ln ∣ x - 2 ∣ - \frac{3}{32} ln ∣ x + 6 ∣ + \frac{3}{32} ln ∣ x - 6 ∣)

Add a constant to the solution

= 64 (\frac{1}{32} ln ∣ x + 2 ∣ - \frac{1}{32} ln ∣ x - 2 ∣ - \frac{3}{32} ln ∣ x + 6 ∣ + \frac{3}{32} ln ∣ x - 6 ∣) + C

What is the integral of (64x^2)/(x^4-40x^2+144) ?
The integral of (64x^2)/(x^4-40x^2+144) is 64(1/32 ln|x+2|-1/32 ln|x-2|-3/32 ln|x+6|+3/32 ln|x-6|)+C