Rewrite the integrand: x^2/x^2 + 4 dx = (x^2 + 4) - 4/x^2 + 4 dx ≤ft( 1 - 4/x^2 + 4 ) dx 1 dx - 4 1/x^2 + 2^2 dx Using dx/x^2+a^2 = 1/a ^-1(x/a) : x - 4 ≤ft[ 1/2 ^-1≤ft( x/2 ) ] + C x - 2 ^-1≤ft( x/2 ) + C

x^2/x^2 + 4 dx = - Mathematics MCQ Mathematics Mock Test - 5 2026 | ExamTest.live

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

A $x - \frac{1}{2} tan^{- 1} (\frac{x}{2}) + c$
B $x - tan^{- 1} (\frac{x}{2}) + c$
C $x - 2 tan^{- 1} (\frac{x}{2}) + c$
D $x + \frac{1}{2} tan^{- 1} (\frac{x}{2}) + c$

Solution & Step-by-step Explanation

Rewrite the integrand: \int \frac{x ^{2}}{x ^{2} + 4} d x = \int \frac{( x ^{2} + 4 ) - 4}{x ^{2} + 4} d x \int (1 - \frac{4}{x ^{2} + 4}) d x \int 1 d x - 4 \int \frac{1}{x ^{2} + 2 ^{2}} d x Using \int \frac{d x}{x ^{2} + a ^{2}} = \frac{1}{a} tan^{- 1} (\frac{x}{a}) : x - 4 [\frac{1}{2} tan^{- 1} (\frac{x}{2})] + C x - 2 tan^{- 1} (\frac{x}{2}) + C

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

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

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

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

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

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