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Magazine Chapter 17: Problem 6
By factorising the denominator and making a substitution find $$ \int \frac{1}{t^{2}-2 t+1} \mathrm{~d} t $$
Short Answer
Expert verified
The integral is \(-\frac{1}{t-1} + C\).
Step by step solution
01
Simplify the Denominator
First, observe the denominator of the integrand, which is given by \( t^2 - 2t + 1 \). This expression is a perfect square trinomial, which can be rewritten as \( (t-1)^2 \). So the integral becomes \( \int \frac{1}{(t-1)^2} \, \mathrm{d}t \).
02
Make the Substitution
To make the integration simpler, perform a substitution. Let \( u = t - 1 \). Then the differential \( \mathrm{d}u = \mathrm{d}t \). Substitute \( u \) and \( \mathrm{d}u \) into the integral, which transforms it to \( \int \frac{1}{u^2} \, \mathrm{d}u \).
03
Integrate with Respect to u
The integral \( \int \frac{1}{u^2} \, \mathrm{d}u \) is a standard integral. Recall that \( \int u^n \, \mathrm{d}u = \frac{u^{n+1}}{n+1} + C \) for \( n eq -1 \). Here \( n = -2 \), so \( \int \frac{1}{u^2} \, \mathrm{d}u = \int u^{-2} \, \mathrm{d}u = \frac{u^{-1}}{-1} + C = -\frac{1}{u} + C \).
04
Substitute Back for t
Now substitute back \( u = t - 1 \) into the result of the integral. This gives \( -\frac{1}{t-1} + C \) as the solution of the original integral.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful technique in integration that simplifies the process by transforming the variable in the integral.
The main idea is to substitute a new variable in place of part of the original expression to make the integral easier to evaluate.
For example, in the integral \( \int \frac{1}{(t-1)^2} \, \mathrm{d}t \), a substitution like \( u = t - 1 \) helps reduce the complexity. Here, the differential \( \mathrm{d}u = \mathrm{d}t \) allows the integral to be rewritten in terms of \( u \), leading to \( \int \frac{1}{u^2} \, \mathrm{d}u \).
The main idea is to substitute a new variable in place of part of the original expression to make the integral easier to evaluate.
For example, in the integral \( \int \frac{1}{(t-1)^2} \, \mathrm{d}t \), a substitution like \( u = t - 1 \) helps reduce the complexity. Here, the differential \( \mathrm{d}u = \mathrm{d}t \) allows the integral to be rewritten in terms of \( u \), leading to \( \int \frac{1}{u^2} \, \mathrm{d}u \).
- Simplifies complex integrals: Reduces calculations by converting the variable.
- Standardizes the integral: Transforms the problem into a more recognizable form.
- Requires derivative matching: Consistency between differentials is key.
Perfect Square Trinomial
Recognizing a perfect square trinomial is crucial to simplify expressions before integrating.
A perfect square trinomial is of the form \( a^2 - 2ab + b^2 = (a-b)^2 \). In our problem, \( t^2 - 2t + 1 \) simplifies to \( (t-1)^2 \).
By rewriting the trinomial, you can transform complex algebraic expressions into a simpler squared form.
A perfect square trinomial is of the form \( a^2 - 2ab + b^2 = (a-b)^2 \). In our problem, \( t^2 - 2t + 1 \) simplifies to \( (t-1)^2 \).
By rewriting the trinomial, you can transform complex algebraic expressions into a simpler squared form.
- Easy recognition: Helps avoid mistakes in simplification.
- Simplifies integration: Transforms expressions into a square, making further steps easier.
- Foundation for substitution: Once simplified, it prepares the expression for substitution.
Definite Integrals
Definite integrals involve integration over a specific interval, providing a specific number as a result.
Though our current exercise focuses on indefinite integration, understanding definite integrals is beneficial.
They are denoted by \( \int_{a}^{b} f(x) \, \mathrm{d}x \), implying that the function \( f(x) \) is integrated over an interval from \( a \) to \( b \).
Though our current exercise focuses on indefinite integration, understanding definite integrals is beneficial.
They are denoted by \( \int_{a}^{b} f(x) \, \mathrm{d}x \), implying that the function \( f(x) \) is integrated over an interval from \( a \) to \( b \).
- Computes area: Determines the area under the curve within the limits.
- Precision: Returns exact numerical values.
- Application: Useful in physics and engineering for determining work done or other quantities across an interval.
Indefinite Integrals
Indefinite integrals, unlike definite integrals, do not evaluate over an interval and are expressed with a constant of integration.
Indefinite integrals find the antiderivative of a function \( f(x) \), denoted by \( \int f(x) \, \mathrm{d}x = F(x) + C \), where \( C \) is the constant of integration.
The result is a family of functions that differ by a constant, providing the general solution to the derivative.
Indefinite integrals find the antiderivative of a function \( f(x) \), denoted by \( \int f(x) \, \mathrm{d}x = F(x) + C \), where \( C \) is the constant of integration.
The result is a family of functions that differ by a constant, providing the general solution to the derivative.
- Determines antiderivatives: Solves for the original function before differentiation.
- Allows general solutions: Provides a solution that encompasses all possible cases.
- Integration constant: Represents any constant that could have been present before differentiation.
