# What is the definite integral of `int_0^8 (1+x^2)/(x^3 +3x) dx`?

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### 2 Answers

The definite integral `int_0^8 (1+x^2)/(x^3 + 3x) dx` has to be determined. First determine the integral `int (1+x^2)/(x^3 + 3x) dx` and find the value of the same between x = 0 and x = 8

`int (1+x^2)/(x^3 + 3x) dx`

Let `x^3 + 3x = y`

`dy/dx = 3x^2 + 3`

`(1 + x^2) dx = (1/3) dy`

`int (1+x^2)/(x^3 + 3x) dx`

= `int (1/3)*(1/y) dy`

= `(1/3)*ln y`

= `(1/3)*ln(x^3 + 3x) + C`

The definite integral `int_0^8 (1+x^2)/(x^3 + 3x) dx`

= `|(1/3)*ln(x^3 + 3x) + C|_0^8`

= `(1/3)*(ln(8^3 + 24) - ln 0)`

But the natural logarithm of 0 is not defined. As a result the given definite integral cannot be determined.

First have the indefinite integral calculation:

`int ((1+x^2)dx)/(x^3+3x)=int((1+x^2)dx)/(x(x^2+3))=int((x^2+3)dx)/(x(x^2+3))-int (2dx)/(x(x^2+3))=`

`int dx/x- 2/3 int(1/x-x/(x^2+3))dx= ` `int dx/x-2/3 int dx/x+2/3 int xdx/(x^2+3)=`

`=1/3 int dx/x+1/3 int 2xdx/(x^2+3)=` `=1/3(logx +log(x^2+3))=log root(3)(x(x^2+3))`

Thus the integral is improprie in the low extreme x=0