Match List I with List II: List I List II A. 1 x + x ? d x I. 2 x + C B. e log x x ? d x II. 2 ( x - 1 ) e x + C C. d x 4 x 2 - 9 III. 2 log ( x + 1 ) + C D. e x ? d x IV. 1 12 log | 2 x - 3 2 x + 3 | + C Choose the correct answer from the op

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Q.
Match List I with List II:

List I List II

A. $\int \frac{1}{x + \sqrt{x}} d x$ I. $2 \sqrt{x} + C$

B. $\int \frac{e^{\log \sqrt{x}}}{x} d x$ II. $2 (\sqrt{x} - 1) e^{\sqrt{x}} + C$

C. $\int \frac{d x}{4 x^{2} - 9}$ III. $2 \log (\sqrt{x} + 1) + C$

D. $\int e^{\sqrt{x}} d x$ IV. $\frac{1}{12} \log$ $| \frac{2 x - 3}{2 x + 3} |$ $+ C$

Choose the correct answer from the option given below:

1 A-II, B-IV, C-I, D-III

2 A-III, B-II, C-IV, D-I

3 A-III, B-I, C-IV, D-II

4 A-I, B-II, C-III, D-IV

Ans.
(3)

Ans. A-III, B-I, C-IV, D-II

Explanation :

$\int \frac{e^{\log \sqrt{x}}}{x} d x = \int \frac{\sqrt{x}}{x} d x = \int \frac{1}{\sqrt{x}} d x = 2 \sqrt{x} + C$

$\int \frac{1}{x + \sqrt{x}} d x = \int \frac{1}{\sqrt{x} (\sqrt{x} + 1)} d x$

$= \int \frac{1}{\sqrt{x}} d x - \int \frac{1}{\sqrt{x} + 1} d x$

$= 2 \sqrt{x} - \log (\sqrt{x} + 1) + C$

Solution

1 A-II, B-IV, C-I, D-III

2 A-III, B-II, C-IV, D-I

3 A-III, B-I, C-IV, D-II

4 A-I, B-II, C-III, D-IV

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	List I		List II
A.	$\int \frac{1}{x + \sqrt{x}} d x$	I.	$2 \sqrt{x} + C$
B.	$\int \frac{e^{\log \sqrt{x}}}{x} d x$	II.	$2 (\sqrt{x} - 1) e^{\sqrt{x}} + C$
C.	$\int \frac{d x}{4 x^{2} - 9}$	III.	$2 \log (\sqrt{x} + 1) + C$
D.	$\int e^{\sqrt{x}} d x$	IV.	$\frac{1}{12} \log$ $\| \frac{2 x - 3}{2 x + 3} \|$ $+ C$

Solution

Q. Match List I with List II: List I List II A. ∫1x+x dx I. 2x+C B. ∫elogxx dx II. 2(x-1)ex+C C. ∫dx4x2-9 III. 2log(x+1)+C D. ∫ex dx IV. 112log|2x-32x+3|+C Choose the correct answer from the option given below:

1 A-II, B-IV, C-I, D-III

2 A-III, B-II, C-IV, D-I

3 A-III, B-I, C-IV, D-II

4 A-I, B-II, C-III, D-IV

Ans. (3) Ans. A-III, B-I, C-IV, D-II Explanation : ∫elogxx dx=∫xx dx=∫1x dx=2x+C ∫1x+x dx=∫1x(x+1) dx =∫1x dx-∫1x+1 dx =2x-log(x+1)+C

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