Chapter 7.5, Problem 46E

a.

To determine

To Find: The integrals of ${\displaystyle \int \frac{dx}{a^2+x^2}}$

Answer to Problem 46E

The integrals of ${\displaystyle \int \frac{dx}{a^2+x^2}}=\frac{\arctan \left(\frac{x}{a}\right)}{a}+c$

Explanation of Solution

Given information: Antiderivatives of various generic functions can be found as formulas in integral tables

Formula used:Antiderivatives of ${\displaystyle \int \frac{dx}{1+x^2}}=\arctan x$

Calculation:

The integrals of ${\displaystyle \int \frac{dx}{a^2+x^2}}$

Take $a^2$ as common

  $\begin{array}{l}{\displaystyle \int \frac{dx}{a^2+x^2}}={\displaystyle \int \frac{dx}{a^2\left(1+{\left(\frac{x}{a}\right)}^2\right)}}\\ \text{let}\\ \frac{x}{a}=t\\ dx=adt\\ {\displaystyle \int \frac{adt}{a^2\left(1+{\left(t\right)}^2\right)}}={\displaystyle \int \frac{dt}{a\left(1+{\left(t\right)}^2\right)}}\\ {\displaystyle \int \frac{dt}{a\left(1+{\left(t\right)}^2\right)}}=\frac{\arctan t}{a}+c\\ \text{but}\\ \text{t=}\frac{x}{a}\\ \frac{\arctan t}{a}+c=\frac{\arctan \left(\frac{x}{a}\right)}{a}+c\\ {\displaystyle \int \frac{dx}{a^2+x^2}}=\frac{\arctan \left(\frac{x}{a}\right)}{a}+c\end{array}$

b.

To determine

To Find: The integrals of ${\displaystyle \int \frac{dx}{a^2-x^2}}$

Answer to Problem 46E

The integrals of ${\displaystyle \int \frac{dx}{a^2-x^2}}=\frac{1}{2a}\left(-\ln \left|a-x\right|+\ln \left|a+x\right|\right)$

Explanation of Solution

Given information: Antiderivatives of various generic functions can be found as formulas in integral tables

Formula used: Antiderivatives of ${\displaystyle \int \frac{dx}{x}}=\ln \left|x\right|$

Calculation:

The integrals of ${\displaystyle \int \frac{dx}{a^2-x^2}}$

  $f(x)=\frac{1}{a^2-x^2}$

The partial fraction decomposition of $f(x)$ is

  $\begin{array}{l}\frac{1}{a^2-x^2}=\frac{1}{\left(a-x\right)\left(a+x\right)}\\ \frac{1}{\left(a-x\right)\left(a+x\right)}=\frac{A}{a-x}+\frac{B}{a+x}\\ 1=A(a+x)+B(a-x)\\ let\\ x=a\\ A=\frac{1}{2a}\\ let\\ x=-a\\ B=\frac{1}{2a}\\ \frac{1}{\left(a-x\right)\left(a+x\right)}=\frac{1}{2a(a-x)}+\frac{1}{2a(a+x)}\\ {\displaystyle \int \frac{dx}{a^2-x^2}}={\displaystyle \int \frac{dx}{\left(a-x\right)\left(a+x\right)}}\\ {\displaystyle \int \frac{dx}{\left(a-x\right)\left(a+x\right)}}=\frac{1}{2a}\left({\displaystyle \int \frac{1}{a-x}dx+{\displaystyle \int \frac{1}{a+x}dx}}\right)\\ \frac{1}{2a}\left({\displaystyle \int \frac{1}{a-x}dx+{\displaystyle \int \frac{1}{a+x}dx}}\right)=\frac{1}{2a}\left(-\ln \left|a-x\right|+\ln \left|a+x\right|\right)\\ {\displaystyle \int \frac{dx}{a^2-x^2}}=\frac{1}{2a}\left(-\ln \left|a-x\right|+\ln \left|a+x\right|\right)\end{array}$

c.

To determine

To Find: The integrals of ${\displaystyle \int \frac{dx}{{\left(a+x\right)}^2}}$

Answer to Problem 46E

The integrals of ${\displaystyle \int \frac{dx}{{\left(a+x\right)}^2}}=\frac{-1}{a+x}+c$

Explanation of Solution

Given information: Antiderivatives of various generic functions can be found as formulas in integral tables

Formula used: Antiderivatives of ${\displaystyle \int \frac{dx}{x^2}}=\frac{-1}{x}$

Calculation:

The integrals of ${\displaystyle \int \frac{dx}{{\left(a+x\right)}^2}}$

  $\begin{array}{l}\text{let}\\ a+x=t\\ dx=dt\\ {\displaystyle \int \frac{dx}{{\left(a+x\right)}^2}}={\displaystyle \int \frac{dt}{{\left(t\right)}^2}}\\ {\displaystyle \int \frac{dt}{{\left(t\right)}^2}}=\frac{-1}{t}+c\\ \text{but}\\ t=a+x\\ \frac{-1}{t}+c=\frac{-1}{a+x}+c\\ {\displaystyle \int \frac{dx}{{\left(a+x\right)}^2}}=\frac{-1}{a+x}+c\end{array}$