∫ t ln(t-2)dt
To evaluate the integral ∫ t ln(t-2) dt, we can use integration by parts. Integration by parts is a technique that allows us to rewrite the integral of a product of two functions as the product of an antiderivative of one function and the integral of another function.
The formula for integration by parts is given by:
∫ u dv = uv - ∫ v du,
where u and v are functions of t and du and dv are their respective differentials.
In this case, we can choose u = ln(t-2) and dv = t dt.
To solve for du and v, we differentiate u and integrate dv:
du = 1/(t-2) dt,
v = ∫ t dt = t^2/2.
Now, we can apply the integration by parts formula:
∫ t ln(t-2) dt = uv - ∫ v du
= ln(t-2) * (t^2/2) - ∫ (t^2/2) * (1/(t-2)) dt.
Simplifying the expression, we have:
∫ t ln(t-2) dt = (t^2/2) ln(t-2) - 1/2 ∫ t^2/(t-2) dt.
Next, we need to evaluate the remaining integral ∫ t^2/(t-2) dt. To do this, we can use partial fraction decomposition.
We can express t^2/(t-2) as a sum of two fractions:
t^2/(t-2) = A + B(t-2),
where A and B are constants to be determined.
By multiplying both sides of the equation by (t-2) and then substituting t=2, we can solve for A:
2^2/(2-2) = A + B(2-2)
4/0 = A + B(0)
A = 4.
Now, let's differentiate both sides of the equation t^2/(t-2) = A + B(t-2) and then substitute t=2. This will allow us to solve for B:
d/dt [t^2/(t-2)] = d/dt [A + B(t-2)]
.....2(t-2)(t-2) = 0 + B(1-0)
.....4 = B.
Therefore, A = 4 and B = 4.
Substituting A = 4 and B = 4 back into the expression for ∫ t^2/(t-2) dt, we obtain:
∫ t ln(t-2) dt = (t^2/2) ln(t-2) - 1/2 ∫ t^2/(t-2) dt
= (t^2/2) ln(t-2) - 1/2 ∫ (4 + 4(t-2)) dt
= (t^2/2) ln(t-2) - 1/2 [4t + 4(t^2/2-2t)] + C,
where C is the constant of integration.
Thus, the final solution is:
∫ t ln(t-2) dt = (t^2/2) ln(t-2) - (2t^2 - 4t) + C.