Chapter 7: Transcendental Functions - Section 7.3 - Exponential Functions - Exercises 7.3 - Page 392: 137

$2\ln5$

$A$ = $\int_{{\,-2}}^{{\,2}}$$\frac{2x}{1+x^{2}}$ $dx$ $A$ = $2\int_{{\,0}}^{{\,2}}$$\frac{2x}{1+x^{2}}$ $dx$ $u$ = $1+x^{2}$ $du$ = $2x$$dx$ $x$ = $0$, $u$ =1 $x$ = $2$, $u$ =5 so $A$ = $2\int_{{\,1}}^{{\,5}}$$\frac{1}{u}$ $du$ $A$ = $2\ln|u|$$|_{{\,1}}^{{\,5}}$ $A$ = $2(\ln5-\ln1)$ $A$ = $2\ln5$