Question
Evaluate: $\int_{-1}^1 \frac{1+x^2}{9-x^2} d x$
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Answer
$ \text { Let } I =\int_{-1}^1 \frac{1+x^2}{9-x^2} d x$
$\therefore I =\int_{-1}^1 \frac{1}{9-x^2} d x+\int_{-1}^1 \frac{x^3}{9-x^2} d x$
$= I _1+ I _2 \ldots \text {...(say) } \ldots . . . \text { (i) } $
Let $f (x)=\frac{1}{9-x^2}$
$ \therefore f (- x )=\frac{1}{9-(-x)^2}$
$=\frac{1}{9-x^2}$
$= f ( x ) $
$\therefore f ( x )$ is an even function.
$ \therefore I _1=\int_{-1}^1 \frac{1}{9-x^2} d x$
$=2 \int_0^1 \frac{1}{9-x^2} d x$
$=2 \int_0^1 \frac{1}{3^2-x^2} d x$
$=2\left[\frac{1}{2 \times 3} \cdot \log \left|\frac{3+x}{3-x}\right|\right]_0^1$
$=\frac{1}{3}\left[\log \left(\frac{4}{2}\right)-\log (1)\right] $
$\therefore I _1=\frac{1}{3} \log 2$
Let $g(x)=\frac{x^3}{9-x^2}$
$ \therefore g(-x)=\frac{(-x)^3}{9-(-x)^2}$
$=\frac{-x^3}{9-x^2}$
$=-g(x) $
$\therefore g(x)$ is an odd function.
$\therefore I _2=\int_{-1}^1 \frac{x^3}{9-x^2} d x=0$
From (i), we get
$ I = I _1+ I _2$
$\therefore I =\frac{1}{3} \log 2+0$
$\therefore I =\frac{1}{3} \log 2 $
$\therefore I =\int_{-1}^1 \frac{1}{9-x^2} d x+\int_{-1}^1 \frac{x^3}{9-x^2} d x$
$= I _1+ I _2 \ldots \text {...(say) } \ldots . . . \text { (i) } $
Let $f (x)=\frac{1}{9-x^2}$
$ \therefore f (- x )=\frac{1}{9-(-x)^2}$
$=\frac{1}{9-x^2}$
$= f ( x ) $
$\therefore f ( x )$ is an even function.
$ \therefore I _1=\int_{-1}^1 \frac{1}{9-x^2} d x$
$=2 \int_0^1 \frac{1}{9-x^2} d x$
$=2 \int_0^1 \frac{1}{3^2-x^2} d x$
$=2\left[\frac{1}{2 \times 3} \cdot \log \left|\frac{3+x}{3-x}\right|\right]_0^1$
$=\frac{1}{3}\left[\log \left(\frac{4}{2}\right)-\log (1)\right] $
$\therefore I _1=\frac{1}{3} \log 2$
Let $g(x)=\frac{x^3}{9-x^2}$
$ \therefore g(-x)=\frac{(-x)^3}{9-(-x)^2}$
$=\frac{-x^3}{9-x^2}$
$=-g(x) $
$\therefore g(x)$ is an odd function.
$\therefore I _2=\int_{-1}^1 \frac{x^3}{9-x^2} d x=0$
From (i), we get
$ I = I _1+ I _2$
$\therefore I =\frac{1}{3} \log 2+0$
$\therefore I =\frac{1}{3} \log 2 $
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