evaluate the following integral.\n∫(90x² / (x⁴ - 53x² + 196)) dx\nfind the partial fraction decomposition of…

Explanation:

Step1: Factor the denominator

Let $t = x^{2}$, then $x^{4}-53x^{2}+196=t^{2}-53t + 196$. Factoring $t^{2}-53t + 196$, we find the roots of the quadratic equation $t^{2}-53t + 196=0$ using the quadratic formula $t=\frac{53\pm\sqrt{53^{2}-4\times196}}{2}=\frac{53\pm\sqrt{2809 - 784}}{2}=\frac{53\pm\sqrt{2025}}{2}=\frac{53\pm45}{2}$. The roots are $t_1 = 49$ and $t_2 = 4$. So $x^{4}-53x^{2}+196=(x^{2}-49)(x^{2}-4)=(x - 7)(x + 7)(x - 2)(x+2)$.

Step2: Set up partial - fraction decomposition

We assume $\frac{90x^{2}}{x^{4}-53x^{2}+196}=\frac{A}{x - 7}+\frac{B}{x + 7}+\frac{C}{x - 2}+\frac{D}{x + 2}$. Then $90x^{2}=A(x + 7)(x - 2)(x + 2)+B(x - 7)(x - 2)(x + 2)+C(x - 7)(x + 7)(x + 2)+D(x - 7)(x + 7)(x - 2)$.

Step3: Find the values of A, B, C, and D

If $x = 7$, then $90\times49=A(7 + 7)(7 - 2)(7 + 2)$, $90\times49=A\times14\times5\times9$, $A = 5$. If $x=-7$, then $90\times49=B(-7 - 7)(-7 - 2)(-7 + 2)$, $90\times49=B\times(- 14)\times(-9)\times(-5)$, $B=-5$. If $x = 2$, then $90\times4=C(2 - 7)(2 + 7)(2 + 2)$, $90\times4=C\times(-5)\times9\times4$, $C=-2$. If $x=-2$, then $90\times4=D(-2 - 7)(-2 + 7)(-2 - 2)$, $90\times4=D\times(-9)\times5\times(-4)$, $D = 2$. So $\frac{90x^{2}}{x^{4}-53x^{2}+196}=\frac{5}{x - 7}-\frac{5}{x + 7}-\frac{2}{x - 2}+\frac{2}{x + 2}$.

Answer:

$\int\frac{90x^{2}}{x^{4}-53x^{2}+196}dx=\int\left(\frac{5}{x - 7}-\frac{5}{x + 7}-\frac{2}{x - 2}+\frac{2}{x + 2}\right)dx$