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

The integral of (1+x^2)/(x(2+x^2)) is 1/4 ln|x^2+2|+1/2 ln(x)+C

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

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

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

Solution steps

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

Expand

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

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

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

= - \frac{1}{4} (ln x^{2} + 2 - 2 ln (x)) + \frac{1}{2} ln 2 + x^{2}

Simplify

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

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

Add a constant to the solution

= \frac{1}{4} ln x^{2} + 2 + \frac{1}{2} ln (x) + C

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What is the integral of (1+x^2)/(x(2+x^2)) ?
The integral of (1+x^2)/(x(2+x^2)) is 1/4 ln|x^2+2|+1/2 ln(x)+C