⇒Indefinite Integration

Correct Option is (C)

$\begin{array}{rl}& \text{ Let }\mathrm{I}=\int \frac{\mathrm{d}x}{x\sqrt{1-x^3}}\\ & \begin{array}{rl}& =\int \frac{x^2\mathrm{d}x}{x^3\sqrt{1-x^3}}\mathrm{d}x\end{array}\\ & \begin{array}{c}\text{ Put }1-x^3={\mathrm{t}}^2\\ \Rightarrow -3x^2\mathrm{d}x=2\mathrm{tdt}\\ \Rightarrow x^2\mathrm{d}x=\frac{-2\mathrm{tdt}}{3}\end{array}\end{array}$

$\begin{array}{rl}\therefore \mathrm{I}& =\int \frac{-\left(\frac{2\mathrm{tdt}}{3}\right)}{(1-{\mathrm{t}}^2)\mathrm{t}}\\ & =-\frac{2}{3}\int \frac{1}{1-{\mathrm{t}}^2}\mathrm{dt}\\ & =\frac{2}{3}\int \frac{1}{{\mathrm{t}}^2-1}\mathrm{dt}\\ & =\frac{2}{3}\times \frac{1}{2}\cdot \log \left|\frac{\mathrm{t}-1}{\mathrm{t}+1}\right|+\mathrm{c}\\ & =\frac{1}{3}\log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+\mathrm{c}\\ \therefore \mathrm{k}& =\frac{1}{3}\end{array}$

Correct Option is (C)

$\begin{array}{rl}\text{ Let }\mathrm{I}& =\int \frac{5\tan x}{\tan x-2}\mathrm{d}x\\ & =\int \frac{5\sin x}{\sin x-2\cos x}\mathrm{d}x\end{array}$

Let $5\sin x=\mathrm{A}(\sin x-2\cos x)+\mathrm{B}\cdot \frac{\mathrm{d}}{\mathrm{d}x}(\sin x-2\cos x)$

$\begin{array}{rl}& \therefore 5\sin x=\mathrm{A}(\sin x-2\cos x)+\mathrm{B}(\cos x+2\sin x)\\ & \therefore \mathrm{A}+2\mathrm{B}=5\text{ and }-2\mathrm{A}+\mathrm{B}=0\end{array}$

Solving these equations, we get

$\begin{array}{rl}\mathrm{A}& =1,\mathrm{B}=2\\ \therefore \mathrm{I}& =\int 1\mathrm{d}x+2\int \frac{\cos x+2\sin x}{\sin x-2\cos x}\mathrm{d}x\\ & =x+2\log |\sin x-2\cos x|+\mathrm{c}\end{array}$

But $\int \frac{5\tan x}{\tan x-2}\mathrm{d}x=x+\mathrm{alog}|\sin x-2\cos x|+\mathrm{c}$

Comparing, we get $\mathrm{a}=2$

Correct Option is (C)

$\begin{array}{rl}\text{ Let }\mathrm{I}& =\int \sqrt{\frac{x-7}{x-9}}\mathrm{d}x\\ & =\int \sqrt{\frac{(x-7)(x-7)}{(x-9)(x-7)}}\mathrm{d}x\\ & =\int \frac{x-7}{\sqrt{x^2-16x+63}}\mathrm{d}x\end{array}$

$\begin{array}{ll}& \text{ Let }(x-7)=A\left[\frac{d}{dx}(x^2-16x+63)\right]+B\\ \therefore & x-7=A(2x-16)+B\\ \therefore & x-7=2Ax-16A+B\\ \therefore & A=\frac{1}{2},B=1\end{array}$

$\begin{array}{rl}& \therefore \mathrm{I}=\int \frac{\frac{1}{2}(2x-16)+1}{\sqrt{x^2-16x+63}}\mathrm{d}x\\ & =\frac{1}{2}\int \frac{2x-16}{\sqrt{x^2-16x+63}}\mathrm{d}x+\int \frac{1}{\sqrt{x^2-16x+63}}\mathrm{d}x\\ & =\frac{1}{2}\times 2\sqrt{x^2-16x+63}+\int \frac{1}{\sqrt{(x-8{)}^2-(1{)}^2}}\mathrm{d}x\\ & \text{... }[\int \frac{{\mathrm{f}}^{\prime }(x)}{\sqrt{\mathrm{f}(x)}}\mathrm{d}x=2\sqrt{\mathrm{f}(x)}+\mathrm{c}]\\ & \therefore \mathrm{I}=\sqrt{x^2-16x+63}+\log |x-8+\sqrt{x^2-16x+63}|+\mathrm{c}\\ & \text{ But, }\int \sqrt{\frac{x-7}{x-9}}\mathrm{d}x=\mathrm{A}\sqrt{x^2-16x+63}\\ & +\log |(x-8)+\sqrt{x^2-16x+63}|+c\end{array}$

Comparing, we get

$\mathrm{A}=1$

Correct Option is (B)

$\begin{array}{rl}\text{ Let I }& =\int \frac{1}{7-6x-x^2}dx\\ & =\int \frac{1}{7-6x-x^2-9+9}dx\\ & =\int \frac{1}{16-(x^2+6x+9)}dx\\ & =\int \frac{1}{(4{)}^2-(x+3{)}^2}dx\\ & =\frac{1}{8}\log \left|\frac{4+x+3}{4-(x+3)}\right|+c\\ & =\frac{1}{8}\log \left|\frac{7+x}{1-x}\right|+c\end{array}$

Correct Option is (D)

$\begin{array}{l}\text{ Let }\mathrm{I}=\int \frac{\sin x}{3+4{\cos }^{2}x}\mathrm{d}x\\ \text{ Put }\cos x=\mathrm{t}\\ -\sin x\mathrm{d}x=\mathrm{dt}\\ \therefore \sin x\mathrm{d}x=-\mathrm{dt}\\ \therefore \mathrm{I}=\int \frac{-\mathrm{dt}}{3+4{\mathrm{t}}^2}\\ =-1\int \frac{1}{4{\mathrm{t}}^2+3}\\ =-1\int \frac{1}{(2\mathrm{t}{)}^2+(\sqrt{3}{)}^2}\\ =-1\times \frac{1}{2\times \sqrt{3}}{\tan }^{-1}\left(\frac{2\mathrm{t}}{\sqrt{3}}\right)+\mathrm{c}\\ \mathrm{I}=\frac{-1}{2\sqrt{3}}{\tan }^{-1}\left(\frac{2\cos x}{\sqrt{3}}\right)+\mathrm{c}\\ \text{ But }\int \frac{\sin x}{3+4{\cos }^{2}x}\mathrm{d}x=\mathrm{A}{\tan }^{-1}(\mathrm{B}\cos x)+\mathrm{c}\\ \text{ Comparing above equations, we get }\\ \text{ A }=\frac{-1}{2\sqrt{3}},\mathrm{B}=\frac{2}{\sqrt{3}}\\ \therefore \text{ A }+\mathrm{B}=\frac{-1}{2\sqrt{3}}+\frac{2}{\sqrt{3}}=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2}\end{array}$