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

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

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integral of 1/(x^2(1+1/x)^3)

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

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

Solution steps

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

Simplify

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

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

Apply u-substitution

: \int \frac{u - 1}{u ^{3}} d u

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

Expand

\frac{u - 1}{u ^{3}} : \frac{1}{u ^{2}} - \frac{1}{u ^{3}}

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

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

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

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

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

Substitute back

u = x + 1

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (2 x^{3} - 3 x^{2} - 36 x + 5)

\int_{0}^{1} 5 x^{4} e^{- x} d x

\int \frac{4}{x ^{2} - 14 x + 49} d x

\frac{\partial}{\partial x} (cos^{2} (5 x))

\frac{d}{d x} (x^{- \frac{3}{7}})

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