If -dx-x2 - 1x2 - 4 - K tan-1 x - L tan-1x-2 - C- then

The correct options are A K = 1/3 D L = 1/6 nbsp;I=∫ dx /( x ^ 2 +1 ) ( x ^ 2 +4 ) nbsp; nbsp; nbsp;= 1 / 3 ∫( 1 /( x ^ 2 +1 ) ...

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Byju's Answer

Standard XII

Mathematics

Integration by Partial Fractions

If ∫dx/x2 +...

Question

If $\int \frac{d x}{(x^{2} + 1) (x^{2} + 4)} = K t a n^{- 1} x + L t a n^{- 1} \frac{x}{2} + C$ , then

K = 1 / 3

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L = 2 / 3

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K = - 1 / 3

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L = - 1 / 6

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Solution

The correct options are
A $K = 1 / 3$
D $L = - 1 / 6$
$I = \int \frac{d x}{(x^{2} + 1) (x^{2} + 4)}$
$= \frac{1}{3} \int (\frac{1}{(x^{2} + 1)} - \frac{1}{(x^{2} + 4)}) d x$
$= \frac{1}{3} {tan}^{- 1} x - \frac{1}{6} {tan}^{- 1} \frac{x}{2} + c$

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\int \frac{d x}{(x^{2} + 1) (x^{2} + 4)} = K t a n^{- 1} x + L t a n^{- 1} \frac{x}{2} + c

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(k + 1) x^{2} - 2 (k - 1) x + 1 = 0

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2 x^{2} + k x + 3 = 0

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If ∫dx(x2+1)(x2+4)=Ktan−1x+Ltan−1x2+C, then

The correct options are A K=1/3 D L=−1/6I=∫dx(x2+1)(x2+4)=13∫(1(x2+1)−1(x2+4))dx=13tan−1x−16tan−1x2+c

If $\int \frac{d x}{(x^{2} + 1) (x^{2} + 4)} = K t a n^{- 1} x + L t a n^{- 1} \frac{x}{2} + C$ , then

The correct options are
A $K = 1 / 3$
D $L = - 1 / 6$
$I = \int \frac{d x}{(x^{2} + 1) (x^{2} + 4)}$
$= \frac{1}{3} \int (\frac{1}{(x^{2} + 1)} - \frac{1}{(x^{2} + 4)}) d x$
$= \frac{1}{3} {tan}^{- 1} x - \frac{1}{6} {tan}^{- 1} \frac{x}{2} + c$