What is the integral of (3-1/x)^{-2}(1/(x^2)) ?

The integral of (3-1/x)^{-2}(1/(x^2)) is -x/((3x-1)^2)+1/(3(3x-1)^2)+C

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integral of (3-1/x)^{-2}(1/(x^2))

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

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

Solution steps

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

Apply Integration By Parts

: - \frac{x}{( 3 x - 1 ) ^{2}} - \int \frac{2}{( 3 x - 1 ) ^{3}} d x

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

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

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

Simplify

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

Add a constant to the solution

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

\int \frac{x ^{3}}{( 5 - x ^{4} ) ^{5}} d x

y = 4 (x - \frac{1}{x})^{3}, at x = 2

\int \frac{1}{( x - 1 ) 4 x ^{2} - 8 x + 3} d x

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

y^{^{''}} - y = 0, y (0) = 0, y^{^{'}} (0) = 1

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