# Calculate antiderivative of `f(x)*sqrt(x+1)?` f`(x)=e^x*sqrt(x+1)`

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You should evaluate the integral of the given function such that:

`int f(x)sqrt(x+1)dx`

You need to substitute `e^xsqrt(x+1)` for `f(x)` such that:

`int f(x)sqrt(x+1)dx = int (e^xsqrt(x+1))sqrt(x+1)dx`

`int f(x)sqrt(x+1)dx = int e^x*(x+1)dx`

`int f(x)sqrt(x+1)dx = int(x*e^x + e^x)dx`

Using the property of linearity of integrals yields:

`int f(x)sqrt(x+1)dx = int(x*e^x)dx + int (e^x)dx`

You need to solve `int(x*e^x)dx ` using parts such that:

`int udv = uv - int vdu`

Reasoning by analogy, yields:

`u = x => du = dx`

`dv = e^xdx => v = e^x`

`int(x*e^x)dx = x*e^x - int e^xdx`

`int f(x)sqrt(x+1)dx = x*e^x - int e^xdx + int e^xdx`

Reducing like integrals yields:

`int f(x)sqrt(x+1)dx = x*e^x + c`

**Hence, evaluating the antiderivative of the given function yields `int f(x)sqrt(x+1)dx = x*e^x + c` .**

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