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

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

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

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

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

Solution steps

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

\int 1 d x = x

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

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

Add a constant to the solution

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

x \to 2 lim (4 - x)

y = \frac{x ^{5}}{3 - x ^{4}}

\frac{\partial}{\partial y} (ln (2 x^{2} + y^{2}))

x \to 4 lim (\frac{x - 4}{x ^{2} + 3 x - 28})

x \to \infty lim (\frac{x}{12})

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