What is the integral of e^{ln(x+1)} ?

The integral of e^{ln(x+1)} is (x^2)/2+x+C

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integral of e^{ln(x+1)}

\int e^{l n (x + 1)} d x

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

Solution steps

\int e^{l n (x + 1)} d x

Apply log rule:

a^{l o g_{a} (b)} = b

= \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

= \int x d x + \int 1 d x

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

\int 1 d x = x

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

Add a constant to the solution

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

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