Question Statement

Q8. ${\int }_2^4\frac{x^2+8}{x^2}dx$

---

Background and Explanation

This problem involves integrating a rational function by first simplifying the integrand into separate terms. You will need to apply the power rule for integration and evaluate definite integrals using the Fundamental Theorem of Calculus.

---

Solution

Begin by simplifying the integrand. Divide each term in the numerator by $x^2$ to separate the fraction:

$\begin{array}{rl}{\int }_2^4\frac{x^2+8}{x^2}dx& ={\int }_2^4\left(\frac{x^2}{x^2}+\frac{8}{x^2}\right)dx\\ & ={\int }_2^4\left(1+\frac{8}{x^2}\right)dx\\ & ={\int }_2^{4}1dx+8{\int }_1^4{x}^{-2}dx\end{array}$

Now apply the power rule for integration to each term. Recall that $\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for $n\ne -1$, and evaluate using the fundamental theorem:

$\begin{array}{rl}{\int }_2^4\frac{x^2+8}{x^2}dx& ={\left[x\right]}_2^4+8{\left[\frac{x^{-2+1}}{-2+1}\right]}_1^4\\ & ={\left[x\right]}_2^4+8{\left[\frac{x^{-1}}{-1}\right]}_1^4\\ & =(4-2)-8{\left[\frac{1}{x}\right]}_1^4\\ & =2-8\left(\frac{1}{4}-\frac{1}{1}\right)\\ & =2-8\left(\frac{1-4}{4}\right)\\ & =2-2(-3)\\ & =2+6\\ & =8\end{array}$

Thus, the value of the definite integral is $\boxed{8}$.

---

Key Formulas or Methods Used

• Algebraic simplification: $\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$
• Power Rule for Integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ (for $n\ne -1$)
• Definite Integral Evaluation: ${\int }_a^{b}f(x)dx=F(b)-F(a)$, where $F$ is the antiderivative of $f$
• Linearity of Integration: $\int (af(x)+bg(x))dx=a\int f(x)dx+b\int g(x)dx$

---

Summary of Steps

1. Split the fraction: Rewrite $\frac{x^2+8}{x^2}$ as $1+8x^{-2}$ (or $1+\frac{8}{x^2}$)
2. Separate the integral: Break into ${\int }_2^{4}1dx+8{\int }_1^4{x}^{-2}dx$
3. Integrate each part: Antiderivative of $1$ is $x$; antiderivative of $x^{-2}$ is $\frac{x^{-1}}{-1}=-\frac{1}{x}$
4. Evaluate at bounds: Calculate $(4-2)-8\left(\frac{1}{4}-1\right)$
5. Simplify arithmetic: $2-8\left(-\frac{3}{4}\right)=2+6=8$