Title: CR_Int_15
Course: Demonstrating STACK (v4.3.7)

Integration using partial fractions

By expressing the integrand as the sum of its partial fractions, find the following integral:
\[ \int [value] d[value]. \]
Feedback: In order to calculate \[\int [value] d[value]\] we need to express \[[value]\] as the sum of its partial fractions alg_partial_fractions. \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value] \\ &\displaystyle =&\displaystyle [value][value] + [value][value] \end{array}\] By multiplying both sides by [value] we get \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value]+[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\] Equating coefficients of [value] we get a system of equations that can be solved to get \(A = [value]\) and \(B = [value]\). So, \[[value] = [value]\] Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \int [value] d[value] + [value] \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] + [value] +c \end{array}\]

By expressing the integrand as the sum of its partial fractions, find the following integral:
\[ \int [value] d[value]. \]
Feedback: In order to calculate \[\int [value] d[value]\] we need to express \[[value]\] as the sum of its partial fractions [[facts:alg_partial_fractions]]. \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value] \\ &\displaystyle =&\displaystyle [value][value] + [value][value] \end{array}\] By multiplying both sides by [value] we get \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value]+[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\] Equating coefficients of [value] we get a system of equations that can be solved to get \(A = [value]\) and \(B = [value]\). So, \[[value] = [value]\] Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \int [value] d[value] + [value] \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] + [value] +c \end{array}\]

By expressing the integrand as the sum of its partial fractions, find the following integral:
\[ \int [value] d[value]. \]
Feedback: In order to calculate \[\int [value] d[value]\] we need to express \[[value]\] as the sum of its partial fractions alg_partial_fractions. \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value] \\ &\displaystyle =&\displaystyle [value][value] + [value][value] \end{array}\] By multiplying both sides by [value] we get \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value]+[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\] Equating coefficients of [value] we get a system of equations that can be solved to get \(A = [value]\) and \(B = [value]\). So, \[[value] = [value]\] Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \int [value] d[value] + [value] \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] + [value] +c \\ &\displaystyle =&\displaystyle [value] +c \end{array}\]

By expressing the integrand as the sum of its partial fractions, find the following definite integral:
\[ \int^[value]_[value] [value] d[value]. \]
Feedback: In order to calculate \[\int [value] d[value]\] we need to express the integrand as the sum of its partial fractions. \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value] \\ &\displaystyle =&\displaystyle [value][value] + [value][value] \end{array}\] By multiplying both sides by [value] we get \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value]+[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\] Equating coefficients we get a system of equations that can be solved to get \(A = [value]\) and \(B = [value]\). So, \[[value] = [value]\] Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \int [value] d[value] + [value] \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] + [value] +c \end{array}\] Then, \[\begin{array}{rcl}\displaystyle \int^[value]_[value] [value] d[value] &\displaystyle =&\displaystyle F([value]) - F([value]) \\ &\displaystyle =&\displaystyle [value] - [[value]] \\ &\displaystyle =&\displaystyle [value]. \end{array}\]

By expressing the integrand as the sum of its partial fractions, find the following definite integral:
\[ \int^[value]_[value] [value] d[value]. \]
Feedback: In order to calculate \[\int [value] d[value]\] we need to express \[[value]\] as the sum of its partial fractions. \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value] \\ &\displaystyle =&\displaystyle [value][value] + [value][value] \end{array}\] By multiplying both sides by [value] we get \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value]+[value] \\ &\displaystyle =&\displaystyle [value] \end{array}\] Equating coefficients we get a system of equations that can be solved to get \(A = [value]\) and \(B = [value]\). So, \[[value] = [value]\] Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \int [value] d[value] + [value] \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] + [value] +c \end{array}\] Recall that if \[\int f(x) dx = F(x)\] then \[\int^a_b f(x) dx = F(a) - F(b).\] Then, \[\begin{array}{rcl}\displaystyle \int^[value]_[value] [value] d[value] &\displaystyle =&\displaystyle F([value]) - F([value]) \\ &\displaystyle =&\displaystyle [value] - [[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

By expressing the integrand as the sum of its partial fractions, find the following definite integral: \[ \int^[value]_[value] [value] d[value]. \]
Feedback: In order to calculate this integral we need to express \[[value]\] as the sum of its partial fractions. Omitting details we have \[[value] = [value],\] hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \int [value] d[value] + [value] \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] + [value] +c \end{array}\] Recall that if \[\int f(x) dx = F(x)\] then \[\int^a_b f(x) dx = F(a) - F(b).\] Then, \[\begin{array}{rcl}\displaystyle \int^[value]_[value] [value] d[value] &\displaystyle =&\displaystyle F([value]) - F([value]) \\ &\displaystyle =&\displaystyle [value] - [[value]] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

By expressing the integrand as the sum of its partial fractions, find the following definite integral: \[ \int^[value]_[value] [value] d[value]. \]
Feedback: In order to calculate \[\int [value] d[value]\] we need to express \[[value]\] as the sum of its partial fractions alg_partial_fractions. Omitting details we have \[[value] = [value].\] Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \int [value] d[value] + [value] \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] + [value] +c \\ &\displaystyle =&\displaystyle [value] +c \end{array}\] We need \[\int^[value]_[value] [value] d[value]\] so, recall that if \[\int f(x) dx = F(x)\] then \[\int^a_b f(x) dx = F(a) - F(b).\] Thus, define \[F([value]) = [value],\] then, \[\begin{array}{rcl}\displaystyle \int^[value]_[value] [value] d[value] &\displaystyle =&\displaystyle F([value]) - F([value]) \\ &\displaystyle =&\displaystyle [value] - \left[[value]\right] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

By expressing the integrand as the sum of its partial fractions, find the following definite integral:
\[ \int^[value]_[value] [value] d[value]. \]
Feedback: In order to calculate \[\int [value] d[value]\] we need to express \[[value]\] as the sum of its partial fractions alg_partial_fractions. \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value] \\ &\displaystyle =&\displaystyle [value][value] + [value][value] \end{array}\] Omitting details we have \[[value] = [value],\] hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \int [value] d[value] + [value] \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] + [value] +c \\ &\displaystyle =&\displaystyle [value] +c \end{array}\] We need \[\int^[value]_[value] [value] d[value]\] so, recall that if \[\int f(x) dx = F(x)\] then \[\int^a_b f(x) dx = F(a) - F(b).\] Thus, define \[F([value]) = [value]\] Then, \[\begin{array}{rcl}\displaystyle \int^[value]_[value] [value] d[value] &\displaystyle =&\displaystyle F([value]) - F([value]) \\ &\displaystyle =&\displaystyle [value] - \left[[value]\right] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

By expressing the integrand as the sum of its partial fractions, find the following definite integral:
\[ \int^[value]_[value] [value] d[value]. \]
Feedback: In order to calculate \[\int [value] d[value]\] we need to express \[[value]\] as the sum of its partial fractions alg_partial_fractions. Omitting details we have \[[value] = [value],\] hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \int [value] d[value] + [value] \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] + [value] +c \\ &\displaystyle =&\displaystyle [value] +c \end{array}\] We need \[\int^[value]_[value] [value] d[value]\] so, recall that if \[\int f(x) dx = F(x)\] then \[\int^a_b f(x) dx = F(a) - F(b).\] Thus, define \[F([value]) = [value]\] Then, \[\begin{array}{rcl}\displaystyle \int^[value]_[value] [value] d[value] &\displaystyle =&\displaystyle F([value]) - F([value]) \\ &\displaystyle =&\displaystyle [value] - \left[[value]\right] \\ &\displaystyle =&\displaystyle [value] \end{array}\]

Find \[ \int [value] d[value]. \]
Feedback: In order to calculate \[\int [value] d[value]\] we need to express \[[value]\] as the sum of its partial fractions alg_partial_fractions. \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value] \\ &\displaystyle =&\displaystyle [value][value] + [value][value] + [value][value] \end{array}\] By multiplying both sides by [value] we get \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value] + [value] + [value] \\ &\displaystyle =&\displaystyle [value] \end{array}\] Equating coefficients of [value] we get a system of equations that can be solved to get \(A = [value]\), \(B = [value]\) and \(C = [value]\). So, \[[value] = [value]\] Hence, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \int [value] d[value] + [value] \int [value] d[value] + [value] \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] + [value] + [value] +c \\ &\displaystyle =&\displaystyle [value] +c \end{array}\]

Find \[ \int [value] d[value]. \] Note that the degree of the numerator is greater then the degree of the denominator.
Feedback: Since the degree of the numerator is greater than the degree of the denominator we need to use long division the get a simpler form to express \[[value].\] alg_partial_fractions Using long division we get that \[[value] = [value] + [value]\] Now, \[[value] = [value][value] + [value][value]\] So, multiplying by [value] both sides we get \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value] + [value] \\ &\displaystyle =&\displaystyle [value] \end{array}\] So, equating cefficients in [value] we get that a system of equations that can be solved to get \(A = [value]\) and \(B = [value]\) Hence, \[[value] = [value]\] So, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \int [value] d[value] + [value] \int [value] d[value] + \int ([value]) d[value] \\ &\displaystyle =&\displaystyle [value] + [value] + [value]+c \\ &\displaystyle =&\displaystyle [value]+c \end{array}\]

Write \[ [value]\] as partial fractions.

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Find \[ \int [value] d[value]. \]
Feedback: Since the degree of the numerator equal to the degree of the denominator we need to use long division. \[[value] = [value] + [value].\] Hence, using partial fractions we have \[[value] = [value].\] So, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \int [value] d[value] + [value] \int [value] d[value] + [value] \int \left([value]\right) d[value] + \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] + [value] + [value]+ [value] + c \\ &\displaystyle =&\displaystyle [value]+c \end{array}\]

Find
\[ \int [value] d[value] \]
by expressing the integrand in partial fractions.
Feedback: Since the degree of the numerator is greater than the degree of the denominator we need to use long division the get a simpler form to express \[[value].\] alg_partial_fractions Using long division we get that \[[value] = [value] + [value]\] Now, \[[value] = [value][value] + [value][value]\] So, multiplying by [value] both sides we get \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value] + [value] \\ &\displaystyle =&\displaystyle [value] \end{array}\] So, equating cefficients in [value] we get that a system of equations that can be solved to get \(A = [value]\) and \(B = [value]\) Hence, \[[value] = [value]\] So, \[\begin{array}{rcl}\displaystyle \int [value] d[value] &\displaystyle =&\displaystyle \int [value] d[value] \\ &\displaystyle =&\displaystyle [value] \int [value] d[value] + [value] \int [value] d[value] + \int ([value]) d[value] \\ &\displaystyle =&\displaystyle [value] + [value] + [value]+c \\ &\displaystyle =&\displaystyle [value]+c \end{array}\]