Match List - I with List - II.
Choose the correct answer from the options given below: |
A - II, B - IV, C - I, D - III A - III, B - II, C - IV, D – I A - III, B - I, C - IV, D - II A - I, B - II, C - III, D - IV |
A - III, B - I, C - IV, D - II |
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 |