What is the integral of (5x^4+9x^2+3)/(x(x^2+1)^2) ?

The integral of (5x^4+9x^2+3)/(x(x^2+1)^2) is ln|x^2+1|-1/(2(x^2+1))+3ln(x)+C

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integral of (5x^4+9x^2+3)/(x(x^2+1)^2)

\int \frac{5 x ^{4} + 9 x ^{2} + 3}{x ( x ^{2} + 1 ) ^{2}} d x

ln x^{2} + 1 - \frac{1}{2 ( x ^{2} + 1 )} + 3 ln (x) + C

Solution steps

\int \frac{5 x ^{4} + 9 x ^{2} + 3}{x ( x ^{2} + 1 ) ^{2}} d x

Expand

\frac{5 x ^{4} + 9 x ^{2} + 3}{x ( x ^{2} + 1 ) ^{2}} : \frac{5 x ^{4}}{x ( x ^{2} + 1 ) ^{2}} + \frac{9 x ^{2}}{x ( x ^{2} + 1 ) ^{2}} + \frac{3}{x ( x ^{2} + 1 ) ^{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{5 x ^{4}}{x ( x ^{2} + 1 ) ^{2}} d x + \int \frac{9 x ^{2}}{x ( x ^{2} + 1 ) ^{2}} d x + \int \frac{3}{x ( x ^{2} + 1 ) ^{2}} d x

\int \frac{5 x ^{4}}{x ( x ^{2} + 1 ) ^{2}} d x = \frac{5}{2} (ln x^{2} + 1 + \frac{1}{x ^{2} + 1})

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

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

= \frac{5}{2} (ln x^{2} + 1 + \frac{1}{x ^{2} + 1}) - \frac{9}{2 ( x ^{2} + 1 )} + 3 ln (x) - \frac{3}{2} ln x^{2} + 1 + \frac{3}{2 ( x ^{2} + 1 )}

Simplify

\frac{5}{2} (ln x^{2} + 1 + \frac{1}{x ^{2} + 1}) - \frac{9}{2 ( x ^{2} + 1 )} + 3 ln (x) - \frac{3}{2} ln x^{2} + 1 + \frac{3}{2 ( x ^{2} + 1 )} : ln x^{2} + 1 - \frac{1}{2 ( x ^{2} + 1 )} + 3 ln (x)

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

Add a constant to the solution

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

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What is the integral of (5x^4+9x^2+3)/(x(x^2+1)^2) ?
The integral of (5x^4+9x^2+3)/(x(x^2+1)^2) is ln|x^2+1|-1/(2(x^2+1))+3ln(x)+C