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

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

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

integral

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

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

Solution steps

\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

y = x^{3} (e^{3})^{4}

\frac{( ( e ^{x} ) ^{2} )}{( 2 x )}

y^{3}

(e^{- x})^{3}

sin^{6} (x)

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