∫ x n - 1 x 2 n + 4 d x =

\int \frac{x^{n - 1}}{x^{2 n} + 4} d x =

$\frac{1}{2 n} \tan^{- 1} (\frac{x^{n}}{2}) + c$
$\frac{n}{2} \tan^{- 1} (\frac{x^{n}}{2}) + c$
$\frac{n}{2} \sin^{- 1} (\frac{x^{n}}{2}) + c$
$\frac{1}{n} \tan^{- 1} (\frac{x^{n}}{2}) + c$

Solution

$\int \frac{x^{n - 1}}{x^{2 n} + 4} d x$

$x^{n} = t$

$\Rightarrow n x^{n - 1} d x = d t$

$= \frac{1}{n} \int \frac{d t}{t^{2} + 4}$

$= \frac{1}{2 n} \tan^{- 1} (\frac{t}{2}) + c$

$= \frac{1}{2 n} \tan^{- 1} (\frac{x^{n}}{2}) + c$

Asked in: AP EAMCET 2020 (23 Sep Shift 1)

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