What is the integral of in(x+1) ?

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

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integral of in(x+1)

\int in (x + 1) d x

in (\frac{x ^{2}}{2} + x) + C

Solution steps

\int in (x + 1) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= in \cdot \int x + 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

= in (\int x d x + \int 1 d x)

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

= in (\frac{x ^{2}}{2} + x)

Add a constant to the solution

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

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(csc (t) + cot (t))^{- 1}