# evaluate the following definite/indefinite integral: | (x^3)(e^x4+4)+(ln(x^2+3)/2x) dx |

### 1 Answer | Add Yours

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

I will change this to three parts and do them separately,

`intx^3e^(x^4) dx + int4x^3 dx + intln(x^2+3)/(2x) dx`

first,`int4x^3 dx`

`int4x^3 dx = x^4`

second,`intx^3e^(x^4) dx`

To do this, you have to make a substitution as x^4 = u.

then,

`4x^3dx = du`

then the integral becomes,

`int e^u* ((du)/4) = e^u/4 = e^(x^4)/4`

Therefore, `intx^3e^(x^4) dx = e^(x^4)/4`

Third part, `intln(x^2+3)/(2x) dx`

To do this also you need to make a substitution.

this time, x^2 = v

then,

`x^2 = v, 2xdx = dv`

`intln(v+3)/(2x) (dv)/(2x)`

`= 1/4intln(v+3)/v dv`

if you analyse this closely you will see that,

`d(ln(v+3)) =(dv)/v`

then the integral becomes,

`= 1/4intln(v+3)*d(ln(v+3))`

this gives,

`=1/4(ln(v+3)^2/2)`

` ` `=ln(v+3)^2/8`

` ` therefore,

`intln(x^2+3)/(2x) dx = ln(x^2+3)^2/8`

Therefore, the total answer is,

`int(x^3(e^(x^4)+4)+ln(x^2+3)/(2x))dx = x^4+e^(x^4)/4+ln(x^2+3)^2/8 +constant`