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Chapter 14: Problem 22

Evaluate the integrals using integration by parts where possible. $$\int\left(t^{2}-t\right) \ln (-t) d t$$

Short Answer

Expert verified
The short answer for evaluating the integral \(\int\left(t^{2}-t\right) \ln (-t) dt\) using integration by parts is given as \(\left(\ln(-t)\right)\left(\frac{1}{3}t^3 - \frac{1}{2}t^2\right) + \frac{1}{9}t^3 - \frac{1}{4}t^2 + C\), where C is the constant of integration.

Step by step solution

01

Choose u and dv

Let u and dv be: \(u = \ln(-t)\) \(dv = (t^2 - t) dt\) Now, we'll differentiate u to find du and integrate dv to find v.
02

Find du and v

Differentiate u with respect to t: \(du = \frac{d}{dt} \ln(-t) dt = \frac{-1}{t} dt\) Integrate dv with respect to t: \(v = \int (t^2 - t) dt = \frac{1}{3}t^3 - \frac{1}{2}t^2 + C_1\) (We can ignore the constant of integration since we'll be subtracting this integral later.)
03

Apply the integration by parts formula

Now that we have u, du, and v, we can apply the integration by parts formula: \(\int\left(t^{2}-t\right) \ln (-t) dt = uv - \int v du\) Substitute our values for u, du, and v and perform the integration: \(\int\left(t^{2}-t\right) \ln (-t) dt = (\ln(-t))\left(\frac{1}{3}t^3 - \frac{1}{2}t^2\right) - \int \left(\frac{1}{3}t^3 - \frac{1}{2}t^2\right)\left(\frac{-1}{t}\right) dt\)
04

Simplify and integrate

Simplify the second integral and integrate with respect to t: \(\int\left(t^{2}-t\right) \ln (-t) dt = \left(\ln(-t)\right)\left(\frac{1}{3}t^3 - \frac{1}{2}t^2\right) + \int \left(\frac{1}{3}t^2 - \frac{1}{2}t\right) dt\) \(\int\left(t^2-t\right) \ln(-t) dt = \left(\ln(-t)\right)\left(\frac{1}{3}t^3 - \frac{1}{2}t^2\right) + \frac{1}{9}t^3 - \frac{1}{4}t^2 + C\) So, the integral \(\int\left(t^{2}-t\right) \ln (-t) dt = \left(\ln(-t)\right)\left(\frac{1}{3}t^3 - \frac{1}{2}t^2\right) + \frac{1}{9}t^3 - \frac{1}{4}t^2 + C\), where C is the constant of integration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indefinite Integral
The concept of an indefinite integral is foundational to calculus and refers to the process of finding the original function from its derivative, commonly referred to as anti-differentiation. Unlike definite integrals, which calculate the area under a curve between two points, an indefinite integral doesn't have limits and results in a family of functions.

To denote an indefinite integral, we use the integral sign followed by the function and a differential, for example, \[ \int f(x) dx \]. The result of an indefinite integral is the original function plus a constant of integration (\

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Most popular questions from this chapter

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