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

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

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

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

x + 1 + \frac{2}{x + 1} + C

Solution steps

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

Complete the square

x^{2} + 2 x - 1 : (x + 1)^{2} - 2

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

Apply u-substitution

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

Expand

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

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

\int 1 d u = u

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

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

Substitute back

u = x + 1

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

Simplify

= x + 1 + \frac{2}{x + 1}

Add a constant to the solution

= x + 1 + \frac{2}{x + 1} + C

f (t) = (1 + t) (8 t^{2} - 5)

n = 1 \sum \infty (1 + \frac{6}{n})^{- n}

\frac{d}{d x} (\frac{4 x}{1 - tan ( x )})

\int \frac{1}{x ^{2} + 60} d x

x \to \infty lim (4 x e^{(\frac{1}{x})} - 4 x)

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