Question

How do you find the definite integral of `int 5/(3x+1)` from [0,4]?

Answer 1

`5/3 ln 13`

Explanation:

To integrate this, one can use substitution method, letting 3x+1 =u. This gives 3dx= du, or dx = `1/3 du`

For x=0, u would be 1 and for x= 4, u would be 13. The given integral thus becomes

`int_(u=1)^13 5/3 1/u du`

= `5/3 [ln u]_1^13`

=`5/3 ln 13`

Answer 2

`5/3ln13`

Explanation:

`"using the standard integral"`

`•color(white)(x)int1/(ax+b)dx=1/aln|ax+b|+c`

`rArrint_0^4 5/(3x+1)dx`

`=[5 . 1/3ln(3x+1)]_0^4`

`=5/3ln13-cancel(5/3ln1)^0`

`=5/3ln13`