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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

\int \frac{y ^{2}}{( y ^{2} + 3 ) ^{\frac{3}{2}}} d y

h \to 0 lim (\frac{sin ( 9 h )}{9 h})

x \to 0 lim (x + 9 - 3)

\frac{d}{d x} (\frac{x ^{2}}{x - 8})

\int \frac{x ^{2}}{x ^{2} + x + 6} d x

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