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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u - 2}{u ^{3}} d u

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

Expand

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

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

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

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

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

Substitute back

u = x + 1

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

Simplify

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

Add a constant to the solution

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

x \to 5 lim (x + 3)

y^{^{'}} + 1 = 2 y

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

\frac{d}{d x} (\frac{a x ^{2}}{ln ( \frac{x}{4} )})

y^{^{'}} = \frac{x - y}{x}

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