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) |
$\frac{1}{12} \log \left|\frac{2 x-3}{2 x+3}\right|+C$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
A. $I=\int \frac{1}{x+\sqrt{x}} d x=\int \frac{1}{\sqrt{x}(\sqrt{x}+1)} d x$
let $y=(\sqrt{x}+1)$ so $d y=\frac{1}{2 \sqrt{x}} d x$
so $2 d y=\frac{d x}{\sqrt{x}}$
$I=\int \frac{2 d y}{y} = 2 \log y + C$
$=2 \log (\sqrt{x}+1) + C$ → III
B. $I=\int \frac{e^{\log \sqrt{x}}}{x} d x$
$=\int \frac{\sqrt{x}}{x} d x$
$=\int x^{-1 / 2} d x$
$=2\sqrt{x} + C$ → I
C. $\int \frac{dx}{4x^2-9} = \frac{1}{4} \int \frac{dx}{x^2-(\frac{3}{2})^2}$
$= \frac{1 × 2}{2 × 4 × 3} \log \left|\frac{x - \frac{3}{2}}{x + \frac{3}{2}} \right|+ C$
$= \frac{1}{12} \log \left|\frac{2x - 3}{2x + 3}\right| + C$ → II
D. $\int e^{\sqrt{x}} d x$
so $y=\sqrt{x} \Rightarrow y^2=x$
so $d x=2 y d y$
$\Rightarrow 2 \int y e^y d y=2\left[y e^y-\int e^y d y\right]$
$=2 y e^y-2 e^y+C$
$=2 e^y(y-1)+C$
$=2 e^{\sqrt{x}}(\sqrt{x}-1)+C$ → II