Match List-I with List-II
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List-I |
List-II |
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A. $\int \frac{d x}{\sqrt{9 x^2-16}}$ |
I. $\frac{1}{3} \sin ^{-1} \frac{3 x}{4}+C$ |
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B. $\int \frac{d x}{\sqrt{16-9 x^2}}$ |
II. $\log \left(e^x+1\right)+C$ |
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C. $\int \frac{e^{2 \log _e x}+1}{e^{2 \log _e x}-1} d x$ |
III. $\frac{1}{3} \log _e\left|3 x+\sqrt{9 x^2-16}\right|+C$ |
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D. $\int \frac{1}{1+e^{-x}} d x$ |
IV. $x+\log _e\left|\frac{x-1}{x+1}\right|+C$ |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 3
A. $\int \frac{d x}{\sqrt{9 x^2-16}} = \frac{1}{3} \int \frac{d x}{\sqrt{x^2-(\frac{4}{3})^2}}$
$\Rightarrow I =\frac{1}{3} \ln \left|x+\sqrt{x^2-\frac{16}{9}}\right|+C$
$\Rightarrow I=\frac{1}{3} \ln \left|3 x+\sqrt{9 x^2-16}\right|+C$ → III
B. $\int \frac{d x}{\sqrt{16-9 x^2}} = \int \frac{d x}{\sqrt{4^2-(3 x)^2}}=\frac{1}{3} \sin ^{-1} \frac{3 x}{4}+C$ → I
C. $\int \frac{e^{2 \log x}+1}{e^{2 \log x}-1} d x = \int \frac{x^2+1}{x^2 - 1}dx$
$= \int \frac{x^2-1}{x^2-1} + \frac{2}{x^2-1}dx = \int 1 + \frac{2}{x^2-1} dx = x+\frac{2}{2} \log _e\left|\frac{x-1}{x+1}\right|+C$ → IV
D. $\int \frac{1}{1+e^{-x}} d x = \int \frac{e^x}{e^x+1} d x = \log \left|e^x+1\right|+C$ → II