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

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

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integral of 1/(x^2(9x^2+1))

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{3}{u ^{2} ( u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{1}{u ^{2} ( u ^{2} + 1 )} d u

Take the partial fraction of

\frac{1}{u ^{2} ( u ^{2} + 1 )} : \frac{1}{u ^{2}} - \frac{1}{u ^{2} + 1}

= 3 \cdot \int \frac{1}{u ^{2}} - \frac{1}{u ^{2} + 1} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

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

= 3 (- \frac{1}{u} - arctan (u))

Substitute back

u = 3 x

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

Simplify

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

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

Add a constant to the solution

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

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