Evaluate the following intregals: ax^2+bx+c (x-a)(x-b)(x-c) dx, where a, b, c are distinct

We have I= ax^2+bx+c (x-a)(x-b)(x-c) dx Let ax^2+bx+c (x-a)(x-b)(x-c) = A x-a + B (x-b) + C (x-c) ^2+bx+c =A(x-b)(x-c)+B(x-c)(x-a)+C(x-a)(x-b ^2+bx+c =A[x^2-(b+c)x+bc]+B[x^2-(c+a)x+ca] +C[x^2-(a-b)x+ab] ^2+]bx+c=(A+B+C)x^2-[A(b+c)+B(c+a) +C(a+b)]x+Abc+Bca+Cab Equation the coefficient on both sides, we get a=A+B+C ...(1) b=-[A(b+c)+B(c+a)+C(a+b)] ...(2) c=Abc+Bca+Cab ...(3) Solving (1), (2), (3) we get A= a^2+ab+c (a-b)(a-c) B= ab^2+b^2+c^2 (b-a)(b-c) C= ac^2+bc+c (c-a)(c-b) = [ a^2+ab+c (a-b)(a-c) 1 x-a + ab^2+b^2+c (b-a)(b-c) 1 x-b + ac^2+bc^2+c (c-a)(c-b) 1 x-c ] dx = a^2+ab+c (a-b)(a-c) |x-a|+ a^2+ab+c (a-b)(a-c) |x-b| + ac^2+bc^2+c (c-a)(c-b) |x-c|+K

Evaluate the following intregals: ax^2+bx+c (x-a)(x-b)(x-c) dx, w…

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Question

Evaluate the following intregals:
$\int\frac{\text{ax}^2+\text{bx}+\text{c}}{(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}-\text{c})}\ \text{dx},$ where a, b, c are distinct

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Answer

We have
$\text{I}=\int\frac{\text{ax}^2+\text{bx}+\text{c}}{(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}-\text{c})}\ \text{dx}$
Let $\int\frac{\text{ax}^2+\text{bx}+\text{c}}{(\text{x}-\text{a})(\text{x}-\text{b})(\text{x}-\text{c})}=\frac{\text{A}}{\text{x}-\text{a}}+\frac{\text{B}}{(\text{x}-\text{b}) }+\frac{ \text{C}}{(\text{x}-\text{c})}$
$\Rightarrow\text{ax}^2+\text{bx}+\text{c}\\=\text{A}(\text{x}-\text{b})(\text{x}-\text{c})+\text{B}(\text{x}-\text{c})(\text{x}-\text{a})+\text{C}(\text{x}-\text{a})(\text{x}-\text{b}$
$\Rightarrow\text{ax}^2+\text{bx}+\text{c}\\=\text{A}[\text{x}^2-(\text{b}+\text{c})\text{x}+\text{bc}]+\text{B}[\text{x}^2-(\text{c}+\text{a})\text{x}+\text{ca}]\\+\text{C}[\text{x}^2-(\text{a}-\text{b})\text{x}+\text{ab}]$
$\Rightarrow\text{ax}^2+]\text{bx}+\text{c}=(\text{A}+\text{B}+\text{C})\text{x}^2-[\text{A}(\text{b}+\text{c})+\text{B}(\text{c}+\text{a}) \\+\text{C}(\text{a}+\text{b})]\text{x}+\text{Abc}+\text{Bca}+\text{Cab}$
Equation the coefficient on both sides, we get
$\text{a}=\text{A}+\text{B}+\text{C}\ ...(1)$
$\text{b}=-[\text{A}(\text{b}+\text{c})+\text{B}(\text{c}+\text{a})+\text{C}(\text{a}+\text{b})]\ ...(2)$
$\text{c}=\text{Abc}+\text{Bca}+\text{Cab}\ ...(3)$
Solving (1), (2), (3) we get
$\text{A}=\frac{\text{a}^2+\text{ab}+\text{c}}{(\text{a}-\text{b})(\text{a}-\text{c})}$
$\text{B}=\frac{\text{ab}^2+\text{b}^2+\text{c}^2}{(\text{b}-\text{a})(\text{b}-\text{c})}$
$\text{C}=\frac{\text{ac}^2+\text{bc}+\text{c}}{(\text{c}-\text{a})(\text{c}-\text{b})}$
$\therefore\text{I}=\int\Big[\frac{\text{a}^2+\text{ab}+\text{c}}{(\text{a}-\text{b})(\text{a}-\text{c})}\times\frac{1}{\text{x}-\text{a}}+\frac{\text{ab}^2+\text{b}^2+\text{c}}{(\text{b}-\text{a})(\text{b}-\text{c})}\times\frac{1}{\text{x}-\text{b}}\\+\frac{\text{ac}^2+\text{bc}^2+\text{c}}{(\text{c}-\text{a})(\text{c}-\text{b})}\times\frac{1}{\text{x}-\text{c}}\Big]\ \text{dx}$
$=\frac{\text{a}^2+\text{ab}+\text{c}}{(\text{a}-\text{b})(\text{a}-\text{c})}\log|\text{x}-\text{a}|+\frac{\text{a}^2+\text{ab}+\text{c}}{(\text{a}-\text{b})(\text{a}-\text{c})}\log|\text{x}-\text{b}|\\+\frac{\text{ac}^2+\text{bc}^2+\text{c}}{(\text{c}-\text{a})(\text{c}-\text{b})}\log|\text{x}-\text{c}|+\text{K}$