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

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

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

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

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

Solution steps

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

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

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (e^{3 x} cos (4 x))

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x^{- 10}

y^{^{'}} + 5 y = 2 e^{(6 \cdot x)}

\frac{\partial}{\partial z} (x^{2} yz - x y z^{3})

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