Match the following ∫f(x) dx is equal to, if Column – I Column – IIA)f(x)=1(x2+1)x2+2(P)x55(1−x4)5/2+CB)f(x)=1(x+2)x2+6x+7(Q)sin−1(x+1(x+2)2)+CC)f(x)=x4+x8(1−x4)7/2(R)(x−2)1−x+cos−1x+CD)f(x)=1−x1+x(S)−tan−11+2x2+C (T)x66(1−x4)5/2+C

A) ∫dx(x2+1)x2+2=∫x13dx(1+1x2)1+2x2 Let 1+2x2=t2C) ∫x4+x8(1−x4)7/2=∫(x+1x3)dx(1x2−x2)7/2 Let 1x2−x2=t2D) Let x=cos2θ⇒dx=−2sin4θ dθ ∫1−x1+x dx=−2∫tanθ sin4θ dθ

Match the following ∫f(x) dx is equal to, if Column – I Column – IIA)f(x)=1(x2+1)x2+2(P)x55(1−x4)5/2+CB)f(x)=1(x+2)x2+6x+7(Q)sin−1(x+1(x+2)2)+CC)f(x)=x4+x8(1−x4)7/2(R)(x−2)1−x+cos−1x+CD)f(x)=1−x1+x(S)−tan−11+2x2+C (T)x66(1−x4)5/2+C

A) ∫dx(x2+1)x2+2=∫x13dx(1+1x2)1+2x2 Let 1+2x2=t2C) ∫x4+x8(1−x4)7/2=∫(x+1x3)dx(1x2−x2)7/2 Let 1x2−x2=t2D) Let x=cos2θ⇒dx=−2sin4θ dθ ∫1−x1+x dx=−2∫tanθ sin4θ dθ

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Match the following $\int f (x) d x$ is equal to, if
Column – I Column – II
A) $f (x) = \frac{1}{(x^{2} + 1) \sqrt{x^{2} + 2}}$ (P) $\frac{x^{5}}{5 {(1 - x^{4})}^{5 / 2}} + C$
B) $f (x) = \frac{1}{(x + 2) \sqrt{x^{2} + 6 x + 7}}$ (Q) $\sin^{- 1} (\frac{x + 1}{(x + 2) \sqrt{2}}) + C$
C) $f (x) = \frac{x^{4} + x^{8}}{{(1 - x^{4})}^{7 / 2}}$ (R) $(\sqrt{x} - 2) \sqrt{1 - x} + \cos^{- 1} \sqrt{x} + C$
D) $f (x) = \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}}$ (S) $- \tan^{- 1} \sqrt{1 + \frac{2}{x^{2}}} + C$
(T) $\frac{x^{6}}{6 {(1 - x^{4})}^{5 / 2}} + C$

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A – S; B – Q; C – P ; D – R

A – R; B – P; C – S; D – R

A – Q; B – R; C – S ; D – S

A – P; B – Q; C – R; D – S

answer is A.

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Detailed Solution

$A) \int \frac{d x}{(x^{2} + 1) \sqrt{x^{2} + 2}} = \int \frac{x^{\frac{1}{3} d x}}{(1 + \frac{1}{x^{2}}) \sqrt{1 + \frac{2}{x^{2}}}} L e t 1 + \frac{2}{x^{2}} = t^{2}$

$C) \int \frac{x^{4} + x^{8}}{{(1 - x^{4})}^{7 / 2}} = \int \frac{(x + \frac{1}{x^{3}}) d x}{{(\frac{1}{x^{2}} - x^{2})}^{7 / 2}} L e t \frac{1}{x^{2}} - x^{2} = t^{2}$

$D) L e t \sqrt{x} = \cos 2 θ \Rightarrow d x = - 2 \sin 4 θ d θ \int \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} d x = - 2 \int \tan θ \sin 4 θ d θ$

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	Column – I		Column – II
A)	$f (x) = \frac{1}{(x^{2} + 1) \sqrt{x^{2} + 2}}$	(P)	$\frac{x^{5}}{5 {(1 - x^{4})}^{5 / 2}} + C$
B)	$f (x) = \frac{1}{(x + 2) \sqrt{x^{2} + 6 x + 7}}$	(Q)	$\sin^{- 1} (\frac{x + 1}{(x + 2) \sqrt{2}}) + C$
C)	$f (x) = \frac{x^{4} + x^{8}}{{(1 - x^{4})}^{7 / 2}}$	(R)	$(\sqrt{x} - 2) \sqrt{1 - x} + \cos^{- 1} \sqrt{x} + C$
D)	$f (x) = \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}}$	(S)	$- \tan^{- 1} \sqrt{1 + \frac{2}{x^{2}}} + C$
		(T)	$\frac{x^{6}}{6 {(1 - x^{4})}^{5 / 2}} + C$

Match the following ∫f(x) dx is equal to, if Column – I Column – IIA)f(x)=1(x2+1)x2+2(P)x55(1−x4)5/2+CB)f(x)=1(x+2)x2+6x+7(Q)sin−1(x+1(x+2)2)+CC)f(x)=x4+x8(1−x4)7/2(R)(x−2)1−x+cos−1x+CD)f(x)=1−x1+x(S)−tan−11+2x2+C (T)x66(1−x4)5/2+C

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