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

\frac{d}{d x} (sin^{- 1} (x))

\int sec^{2} (X) d X

x \to - 3 - lim (\frac{3}{x ^{2} - 9})

\int (3 e^{x} + 5 sin (x)) d x

\int_{1}^{2} x sin (x^{2}) d x

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