The integral from 1 to 5 of 1/x^2*(times) the square root of 1+(1/x)dx. i've had tried this problem over and over again and cannot get the correct answer. Please help. thanks
To solve this integral, we'll use a substitution. Let's set u = 1 + (1/x). Then, we can find du/dx as follows:
du/dx = d/dx (1 + 1/x)
= 0 + d/dx (1/x)
= -1/x^2
Next, we'll solve for dx in terms of du by rearranging the equation above:
dx = -x^2 * du
Now, we can rewrite the integral using our substitution:
∫(1 to 5) (1/x^2) * √(1 + (1/x)) dx = ∫(1 to 5) (1/x^2) * √u * (-x^2 du)
Simplifying, we get:
∫(1 to 5) -√u du
Now, we can integrate as follows:
∫ -√u du = -2/3 * u^(3/2) + C
Substituting back in for u and evaluating the definite integral from 1 to 5, we get:
-2/3 * (1 + (1/5))^(3/2) - (-2/3 * (1 + 1)^3/2)
Simplifying further, we find the answer to the integral is:
-2/3 * (26/25)^(3/2) + 2/3