What is the integral of (x^2)/(x^4+10x^2+9) ?

The integral of (x^2)/(x^4+10x^2+9) is -1/8 arctan(x)+3/8 arctan(x/3)+C

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integral of (x^2)/(x^4+10x^2+9)

\int \frac{x ^{2}}{x ^{4} + 10 x ^{2} + 9} d x

- \frac{1}{8} arctan (x) + \frac{3}{8} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{x ^{2}}{x ^{4} + 10 x ^{2} + 9} d x

Factor

\frac{x ^{2}}{x ^{4} + 10 x ^{2} + 9}

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

Take the partial fraction of

\frac{x ^{2}}{( x ^{2} + 1 ) ( x ^{2} + 9 )} : - \frac{1}{8 ( x ^{2} + 1 )} + \frac{9}{8 ( x ^{2} + 9 )}

= \int - \frac{1}{8 ( x ^{2} + 1 )} + \frac{9}{8 ( x ^{2} + 9 )} 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} + 1 )} d x + \int \frac{9}{8 ( x ^{2} + 9 )} d x

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

\int \frac{9}{8 ( x ^{2} + 9 )} d x = \frac{3}{8} arctan (\frac{x}{3})

= - \frac{1}{8} arctan (x) + \frac{3}{8} arctan (\frac{x}{3})

Add a constant to the solution

= - \frac{1}{8} arctan (x) + \frac{3}{8} arctan (\frac{x}{3}) + C

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- 2 x} ln (x)

f (x) = \frac{1}{7 - x}

\frac{d}{d x} (h (x))

f (x) = \frac{9}{x}

\frac{d y}{d x} = \frac{1}{6} y cos^{2} (y)

What is the integral of (x^2)/(x^4+10x^2+9) ?
The integral of (x^2)/(x^4+10x^2+9) is -1/8 arctan(x)+3/8 arctan(x/3)+C