What is the integral of (x^4+81)/(x(x^2+9)^2) ?

The integral of (x^4+81)/(x(x^2+9)^2) is 9/(x^2+9)+ln(x)+C

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integral of (x^4+81)/(x(x^2+9)^2)

\int \frac{x ^{4} + 81}{x ( x ^{2} + 9 ) ^{2}} d x

\frac{9}{x ^{2} + 9} + ln (x) + C

Solution steps

\int \frac{x ^{4} + 81}{x ( x ^{2} + 9 ) ^{2}} d x

Expand

\frac{x ^{4} + 81}{x ( x ^{2} + 9 ) ^{2}} : \frac{x ^{4}}{x ( x ^{2} + 9 ) ^{2}} + \frac{81}{x ( x ^{2} + 9 ) ^{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{x ^{4}}{x ( x ^{2} + 9 ) ^{2}} d x + \int \frac{81}{x ( x ^{2} + 9 ) ^{2}} d x

\int \frac{x ^{4}}{x ( x ^{2} + 9 ) ^{2}} d x = \frac{1}{2} (ln x^{2} + 9 + \frac{9}{x ^{2} + 9})

\int \frac{81}{x ( x ^{2} + 9 ) ^{2}} d x = ln (x) - \frac{1}{2} ln x^{2} + 9 + \frac{9}{2 ( x ^{2} + 9 )}

= \frac{1}{2} (ln x^{2} + 9 + \frac{9}{x ^{2} + 9}) + ln (x) - \frac{1}{2} ln x^{2} + 9 + \frac{9}{2 ( x ^{2} + 9 )}

Simplify

\frac{1}{2} (ln x^{2} + 9 + \frac{9}{x ^{2} + 9}) + ln (x) - \frac{1}{2} ln x^{2} + 9 + \frac{9}{2 ( x ^{2} + 9 )} : \frac{9}{x ^{2} + 9} + ln (x)

= \frac{9}{x ^{2} + 9} + ln (x)

Add a constant to the solution

= \frac{9}{x ^{2} + 9} + ln (x) + C

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What is the integral of (x^4+81)/(x(x^2+9)^2) ?
The integral of (x^4+81)/(x(x^2+9)^2) is 9/(x^2+9)+ln(x)+C