What is the integral of (1+2x^2)/(x^4(1+x^2)) ?

The integral of (1+2x^2)/(x^4(1+x^2)) is -1/x-1/(3x^3)-arctan(x)+C

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integral of (1+2x^2)/(x^4(1+x^2))

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

- \frac{1}{x} - \frac{1}{3 x ^{3}} - arctan (x) + C

Solution steps

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

Expand

\frac{1 + 2 x ^{2}}{x ^{4} ( 1 + x ^{2} )} : \frac{1}{x ^{4} ( 1 + x ^{2} )} + \frac{2 x ^{2}}{x ^{4} ( 1 + x ^{2} )}

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

\int \frac{1}{x ^{4} ( 1 + x ^{2} )} d x = \frac{1}{x} - \frac{1}{3 x ^{3}} + arctan (x)

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

= \frac{1}{x} - \frac{1}{3 x ^{3}} + arctan (x) + 2 (- \frac{1}{x} - arctan (x))

Simplify

\frac{1}{x} - \frac{1}{3 x ^{3}} + arctan (x) + 2 (- \frac{1}{x} - arctan (x)) : - \frac{1}{x} - \frac{1}{3 x ^{3}} - arctan (x)

= - \frac{1}{x} - \frac{1}{3 x ^{3}} - arctan (x)

Add a constant to the solution

= - \frac{1}{x} - \frac{1}{3 x ^{3}} - arctan (x) + C

What is the integral of (1+2x^2)/(x^4(1+x^2)) ?
The integral of (1+2x^2)/(x^4(1+x^2)) is -1/x-1/(3x^3)-arctan(x)+C