Question

Evaluate
${\int }_{-1}^1\frac{1+x^3}{9-x^2}dx$

Answer 1

$I={\int }_{-1}^1\frac{1+x^3}{9-x^2}dx$ ..........(i)
$I=\int \frac{1+(0-x{)}^3}{9-(0-x{)}^2}dx$
$I=\int \frac{1-x^3}{9-x^2}dx$ ............(ii)
(i) + (ii)
$2I={\int }_{-1}^1\frac{2}{9-x^2}dx$
$I={\int }_{-1}^1\frac{dx}{(3{)}^2-(x{)}^2}$
$=\frac{1}{2\times 3}{\left[\log \frac{5+x}{3-x}\right]}_{-1}^1$
$=\frac{1}{6}(\log \frac{3+1}{3-1}-\log \frac{3-1}{3-(-1)})$
$=\frac{1}{6}(\log \frac{4}{2}-\log \frac{2}{4})$
$=\frac{1}{6}(\log 2-\log 1/2)$
$=\frac{1}{6}\left(\log 4\right)=\frac{2}{6}\log 2=\frac{\log 2}{3}$