# What is #f(x) = int (x-2)/((x+1)(x-4) ) dx# if #f(2)=5 #?

Use integration by partial fractions.

We can hence write the following system of equations.

Solve:

Hopefully this helps!

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The function f(x) is equal to the integral of (x-2)/((x+1)(x-4)) dx. Given that f(2) = 5, we need to find the value of the integral at x = 2.

Using partial fraction decomposition, the integrand can be rewritten as:

(x-2)/((x+1)(x-4)) = A/(x+1) + B/(x-4)

By equating coefficients, we find that A = 3/5 and B = -1/5.

Now, we integrate each term separately:

∫(3/5)/(x+1) dx + ∫(-1/5)/(x-4) dx

= (3/5)ln|x+1| - (1/5)ln|x-4| + C

Now we evaluate f(2) by substituting x = 2:

f(2) = (3/5)ln|2+1| - (1/5)ln|2-4|

= (3/5)ln(3) - (1/5)ln(2)

Given f(2) = 5, we solve for C:

5 = (3/5)ln(3) - (1/5)ln(2) + C

C = 5 - (3/5)ln(3) + (1/5)ln(2)

So, the function f(x) = (3/5)ln|x+1| - (1/5)ln|x-4| + 5 - (3/5)ln(3) + (1/5)ln(2).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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