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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (4 x + 8 x^{\frac{3}{8}})

\frac{d}{d x} \frac{( x ^{2} - 4 ) ^{3}}{36}

x \to 1 + lim (\frac{1}{x ^{3}})

(- 5, - 1), (0, 3)

\frac{d y}{d x} = \frac{y + 1}{x ^{3}}

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