MCQ
$\int_{-1}^1 \frac{1+x^3}{9-x^2} d x=$
  • A
    $\log 2$
  • $\frac{1}{3} \log 2$
  • C
    $\log 9$
  • D
    $\frac{1}{3} \log 9$

Answer

Correct option: B.
$\frac{1}{3} \log 2$
(B)
$\int_{-1}^1 \frac{1+x^3}{9-x^2} d x=\int_{-1}^1 \frac{1}{9-x^2} d x+\int_{-1}^1 \frac{x^3}{9-x^2} d x$
$=2 \int_0^1 \frac{1}{9-x^2} d x+0$
$\ldots\left[\begin{array}{c}\because \frac{1}{9-x^2} \text { is an even function and } \\ \frac{x^3}{9-x^2} \text { is an odd function. }\end{array}\right]$
$=2\left[\frac{1}{2 \times 3} \log \left|\frac{3+x}{3-x}\right|\right]_0^1=\frac{1}{3}(\log 2-\log 1)=\frac{1}{3} \log 2$

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