# Determine the definite integral for y=x+1/x. x=1 and x=3

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We have to find the definite integral for y = x + 1/x between x=1 and x=3.

Now y = x + 1/x

The integral of this is Int [x + 1/x]

=> Int [x] + Int [1/x]

=> x^2 / 2 + ln x + C

For x = 3, x^2 / 2 + ln x + C = 9/2 + ln 3 + C

For x = 1, x^2 / 2 + ln x + C = 1/2 + ln 1 + C

Subtracting what we have found above, the definite integral of y = x + 1/x, between x = 1 and x = 3 is

9/2 + ln 3 + C - 1/2 - ln 1 - C

=> 8/2 + ln 3 - 0

=> 4 + ln 3.

**The required result is 4 + ln 3.**

The definite integral is calculated using Leibniz-Newton formula.

Int (x + 1/x)dx = F(b) - F(a), where a = 1 and b = 3

First, we'll determine the result of the indefinite integral:

Int (x + 1/x)dx

We'll use the additive property of integrals:

Int (x + 1/x)dx = Int xdx + Int dx/x

Int (x + 1/x)dx = x^2/2 + ln x + C

The resulted expression is F(x).

Now, we'll determine F(b) = F(3) and F(a) = F(1):

F(3) = 3^2/2 + ln 3

F(3) = 9/2 + ln 3

F(1) = 1/2 + ln 1

F(1) = 1/2 + 0

F(1) = 1/2

We'll determine the definite integral:

Int (x + 1/x)dx = F(3) - F(1)

Int (x + 1/x)dx = 9/2 + ln 3 - 1/2

We'll combine like terms:

**Int (x + 1/x)dx = 4 + ln 3**