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

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

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

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

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

Solution steps

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

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

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

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

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

\int 1 d x = x

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

Simplify

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

Add a constant to the solution

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

x \to 1 lim (4 x^{3})

x \to - 6 lim (5)

\frac{d}{d x} (16 x^{3} - 96 x^{2} + 178 x - 95)

\frac{e ^{x} x - e ^{x}}{7 x ^{2}}

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