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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

y = 9 x^{3} - x^{2} + 5 x - 6

\int e^{0.02 x} d x

\frac{d}{d x} (4 x ln (x))

\frac{d}{d x} (- cos (x) ln (sec (x) + tan (x)))

x^{3} - 6 x^{2} + 9 x + 1, x = 0.5, x = 4

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