What is the integral of (40x^2)/(x^4-52x^2+576) ?

The integral of (40x^2)/(x^4-52x^2+576) is 40(1/10 ln|x+4|-1/10 ln|x-4|-3/20 ln|x+6|+3/20 ln|x-6|)+C

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integral of (40x^2)/(x^4-52x^2+576)

\int \frac{40 x ^{2}}{x ^{4} - 52 x ^{2} + 576} d x

40 (\frac{1}{10} ln ∣ x + 4 ∣ - \frac{1}{10} ln ∣ x - 4 ∣ - \frac{3}{20} ln ∣ x + 6 ∣ + \frac{3}{20} ln ∣ x - 6 ∣) + C

Solution steps

\int \frac{40 x ^{2}}{x ^{4} - 52 x ^{2} + 576} d x

Take the constant out:

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

= 40 \cdot \int \frac{x ^{2}}{x ^{4} - 52 x ^{2} + 576} d x

Take the partial fraction of

\frac{x ^{2}}{x ^{4} - 52 x ^{2} + 576} : \frac{1}{10 ( x + 4 )} - \frac{1}{10 ( x - 4 )} - \frac{3}{20 ( x + 6 )} + \frac{3}{20 ( x - 6 )}

= 40 \cdot \int \frac{1}{10 ( x + 4 )} - \frac{1}{10 ( x - 4 )} - \frac{3}{20 ( x + 6 )} + \frac{3}{20 ( 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

= 40 (\int \frac{1}{10 ( x + 4 )} d x - \int \frac{1}{10 ( x - 4 )} d x - \int \frac{3}{20 ( x + 6 )} d x + \int \frac{3}{20 ( x - 6 )} d x)

\int \frac{1}{10 ( x + 4 )} d x = \frac{1}{10} ln ∣ x + 4 ∣

\int \frac{1}{10 ( x - 4 )} d x = \frac{1}{10} ln ∣ x - 4 ∣

\int \frac{3}{20 ( x + 6 )} d x = \frac{3}{20} ln ∣ x + 6 ∣

\int \frac{3}{20 ( x - 6 )} d x = \frac{3}{20} ln ∣ x - 6 ∣

= 40 (\frac{1}{10} ln ∣ x + 4 ∣ - \frac{1}{10} ln ∣ x - 4 ∣ - \frac{3}{20} ln ∣ x + 6 ∣ + \frac{3}{20} ln ∣ x - 6 ∣)

Add a constant to the solution

= 40 (\frac{1}{10} ln ∣ x + 4 ∣ - \frac{1}{10} ln ∣ x - 4 ∣ - \frac{3}{20} ln ∣ x + 6 ∣ + \frac{3}{20} ln ∣ x - 6 ∣) + C

(2, 3), (- 2, - 3)

\frac{d}{d x} (\frac{4}{2 x + 1})

f (x) = 2 x^{2} + x - 1, at x = 0

\frac{d}{d x} (lo g_{10} (\frac{x}{x ^{2} + 1}))

\frac{\partial}{\partial x} (15 - 2 x^{2} - y^{2})

What is the integral of (40x^2)/(x^4-52x^2+576) ?
The integral of (40x^2)/(x^4-52x^2+576) is 40(1/10 ln|x+4|-1/10 ln|x-4|-3/20 ln|x+6|+3/20 ln|x-6|)+C