What is the integral of (x^2)/(x^4-42x^2-343) ?

The integral of (x^2)/(x^4-42x^2-343) is 1/(8sqrt(7))arctan(x/(sqrt(7)))-1/16 ln|x+7|+1/16 ln|x-7|+C

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integral of (x^2)/(x^4-42x^2-343)

\int \frac{x ^{2}}{x ^{4} - 42 x ^{2} - 343} d x

\frac{1}{8 7} arctan (\frac{x}{7}) - \frac{1}{16} ln ∣ x + 7 ∣ + \frac{1}{16} ln ∣ x - 7 ∣ + C

Solution steps

\int \frac{x ^{2}}{x ^{4} - 42 x ^{2} - 343} d x

Take the partial fraction of

\frac{x ^{2}}{x ^{4} - 42 x ^{2} - 343} : \frac{1}{8 ( x ^{2} + 7 )} - \frac{1}{16 ( x + 7 )} + \frac{1}{16 ( x - 7 )}

= \int \frac{1}{8 ( x ^{2} + 7 )} - \frac{1}{16 ( x + 7 )} + \frac{1}{16 ( x - 7 )} 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}{8 ( x ^{2} + 7 )} d x - \int \frac{1}{16 ( x + 7 )} d x + \int \frac{1}{16 ( x - 7 )} d x

\int \frac{1}{8 ( x ^{2} + 7 )} d x = \frac{1}{8 7} arctan (\frac{x}{7})

\int \frac{1}{16 ( x + 7 )} d x = \frac{1}{16} ln ∣ x + 7 ∣

\int \frac{1}{16 ( x - 7 )} d x = \frac{1}{16} ln ∣ x - 7 ∣

= \frac{1}{8 7} arctan (\frac{x}{7}) - \frac{1}{16} ln ∣ x + 7 ∣ + \frac{1}{16} ln ∣ x - 7 ∣

Add a constant to the solution

= \frac{1}{8 7} arctan (\frac{x}{7}) - \frac{1}{16} ln ∣ x + 7 ∣ + \frac{1}{16} ln ∣ x - 7 ∣ + C

\frac{4 cos ( x )}{1 + sin ( x )}

y = 8 x e^{x}, (0, 0)

\int_{0}^{\frac{π}{10}} cos (5 x) d x

\int tan^{3} (7 x) sec (7 x) d x

- 2 x^{3}

What is the integral of (x^2)/(x^4-42x^2-343) ?
The integral of (x^2)/(x^4-42x^2-343) is 1/(8sqrt(7))arctan(x/(sqrt(7)))-1/16 ln|x+7|+1/16 ln|x-7|+C