If∫(x+2)(x2+1)dx=aln(1+x2)+btan−1x+51ln∣x+2∣+C Then a=101,b=−52 (b) a=101,b=−52 a=−101,b=52 (d) a=101,b=52

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Question

$I f \int \frac{d x}{( x + 2 ) ( x ^{2} + 1 )} = a ln (1 + x^{2}) + b tan^{- 1} x + \frac{1}{5} ln ∣ x + 2 ∣ + C$ Then $a = \frac{1}{10}, b = - \frac{2}{5}$ (b) $a = \frac{1}{10}, b = - \frac{2}{5}$ $a = - \frac{1}{10}, b = \frac{2}{5}$ (d) $a = \frac{1}{10}, b = \frac{2}{5}$

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I f \int \frac{d x}{( x + 2 ) ( x ^{2} + 1 )} = a ln (1 + x^{2}) + b tan^{- 1} x + \frac{1}{5} ln ∣ x + 2 ∣ + C

Then

a = \frac{1}{10}, b = - \frac{2}{5}

(b)

a = \frac{1}{10}, b = - \frac{2}{5}

a = - \frac{1}{10}, b = \frac{2}{5}

(d)

a = \frac{1}{10}, b = \frac{2}{5}

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Question Text

I f \int \frac{d x}{( x + 2 ) ( x ^{2} + 1 )} = a ln (1 + x^{2}) + b tan^{- 1} x + \frac{1}{5} ln ∣ x + 2 ∣ + C

Then

a = \frac{1}{10}, b = - \frac{2}{5}

(b)

a = \frac{1}{10}, b = - \frac{2}{5}

a = - \frac{1}{10}, b = \frac{2}{5}

(d)

a = \frac{1}{10}, b = \frac{2}{5}