# evaluate the following definite/indefinite integral: || x^3e^x+4+(ln(x^2+3)/2x) dx

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### 1 Answer

Since the problem does not provide the limits of integration, hence you need to evaluate the indefinite integral such that:

`int (x^3e^x+4+ ln((x^2+3)/(2x))) dx`

You need to split the integral into three simpler integrals to make the process of evaluation easier such that:

`int (x^3e^x+4+ ln((x^2+3)/(2x))) dx =int (x^3e^x)dx + int 4dx + int ln((x^2+3)/(2x)) dx`

You need to evaluate the first integral using the formula of integration by parts such that:

`int udv = uv - int vdu`

You should come up with the substitution such that:

`u = x^3 =gt du = 3x^2 dx`

`dv = e^x dx =gt v = e^x`

`int (x^3e^x)dx = (x^3e^x) - int (3x^2e^x)dx`

`u = x^2 =gt du = 2xdx`

`v = e^x`

`int (3x^2e^x)dx = 3e^x*x^2 - 6int xe^x dx`

`u = x =gt du = dx`

`v = e^x`

`6int xe^x dx = 6xe^x - 6int e^x`

`6int xe^x dx = 6xe^x - 6 e^x`

`int (3x^2e^x)dx = 3e^x*x^2 - 6xe^x+ 6 e^x`

`int (x^3e^x)dx = x^3*e^x - 3e^x*x^2+ 6xe^x- 6 e^x + c`

You need to evaluate the second integral such that:

`int 4dx = 4x + c`

You need to evaluate the third integral using integration by parts method such that:

`int ln((x^2+3)/(2x)) dx`

`u = ln ((x^2+3)/(2x)) =gt du = (2x)/(x^2+3)*(4x^2 - 2x^2- 6)/(4x^2) dx`

`du = (2(x^2-3))/(x(x^2+3))`

`dv = dx =gt v = x`

`int ln((x^2+3)/(2x)) dx = x*ln (x^2+3)/(2x) - int (2(x^2-3))/(x^2+3)dx`

You need to evaluate `int (2(x^2-3))/(x^2+3)` such that:

`2int (x^2-3)/(x^2+3)dx = 2int (x^2 + 3 - 6)/(x^2+3)dx`

`2int (x^2 + 3 - 6)/(x^2+3)dx = 2int (x^2 + 3)/(x^2+3)dx - 12int (dx)/(x^2+3)`

`2int (x^2 + 3 - 6)/(x^2+3)dx = 2x - 12/sqrt3*arctan (x/sqrt3) + c`

**Hence, evaluating the given integral yields `int (x^3e^x+4+ ln((x^2+3)/(2x)))dx = x^3*e^x - 3e^x*x^2 + 6xe^x - 6 e^x + 4x + 2x - 12/sqrt3*arctan (x/sqrt3) + c` .**