Integration of $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$ with respect to $x$ :
Answer
Let $I =\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) d x$
$=\int\left(x^{\frac{1}{2}}+x^{-\frac{1}{2}}\right) d x$
$=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C$
$=\frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+C$
Hence correct option is $(C).$
Integration of function $\frac{x}{e^{x^2}}$ with respect to $x$ is equal to :
Answer
(D) $\int \frac{x}{e^{x^2}} d x$ Let $x^2=t \quad \therefore 2 x d x=d t$ $\Rightarrow \quad x d x=\frac{1}{2} d t$ $ \begin{aligned} \int \frac{\frac{1}{2} d t}{e^t} & =\frac{1}{2} \int e^{-t} d t \\ & =\frac{-1}{2} e^{-t}+C \\ & =\frac{-1}{2 e^t}+C=\frac{-1}{2 e^{x^2}}+C \end{aligned} $ Hence correct option is (D).
$\int_0^1 \log \left(\frac{1}{x}-1\right) d x$ is equal to :
Answer
(A)Let $ \begin{aligned} I & =\int_0^1 \log \left(\frac{1}{x}-1\right) d x=\int_0^1 \log \left(\frac{1-x}{x}\right) d x \ldots(1) \\ I & =\int_0^1 \log \left(\frac{1-x}{x}\right) d x \\ & =\int_0^1 \log \left(\frac{1-(1-x)}{(1-x)}\right) d x \text { From Property } P_5 \end{aligned} $ $ \begin{aligned} I & =\int_0^1 \log \left(\frac{1-1+x}{(1-x)}\right) d x=\int_0^1 \log \left(\frac{x}{(1-x)}\right) d x \\ I & =-\int_0^1 \log \left(\frac{1-x}{x}\right) d x \\ I & =-I \\ 2 I & =0 \therefore I=0 \end{aligned} $ Hence option (A) is correct.