Question

Match the statements of Column I with values of Column II
$\begin{array}{cc}ColumnI& ColumnII\\ \left(A\right)\int \frac{e^{2x}-2e^x}{e^{2x}+1}dx=Aln(e^{2x}+1)+Bta{n}^{-1}\left(e^x\right)+c& \left(p\right)A=-\frac{1}{2},B=-\frac{1}{4}\\ \left(B\right)\int \sqrt{x+\sqrt{x^2+2}}dx=A\{x+\sqrt{x^2+2}{\}}^{\frac{3}{2}}+\frac{B}{\sqrt{x+\sqrt{x^2+2}}}+c& \left(q\right)A=\frac{1}{2},B=-2\\ \left(C\right)\int \frac{cos8x-cos7x}{1+2cos5x}dx=Asin3x+Bsin2x+c& \left(r\right)A=\frac{1}{3},B=-2\\ \left(D\right)\int \frac{lnx}{x^3}dx=A\frac{lnx}{x^2}+\frac{B}{x^2}+c& \left(s\right)A=\frac{1}{3},B=-\frac{1}{2}\end{array}$

• A. A-s , B-r , C-q , D-p
• B. A-p , B-s , C-r , D-q
• C. A-p , B-q , C-r , D-s
• D. A-q , B-r , C-s , D-p

Answer 1

The correct option is D

A-q , B-r , C-s , D-p

(A) $\int \frac{e^{2x}}{e^{2x}+1}dx-2\int \frac{e^x}{e^{2x}+1}dx=\frac{1}{2}ln(1+e^{2x})-2ta{n}^{-1}\left(e^x\right)+c$

(B) $x+\sqrt{x^2+2}=t$
$\Rightarrow x^2+2=t^2+x^2-2tx$
$\Rightarrow x=\frac{1}{2}(t-\frac{2}{t})$
So, $I=\frac{1}{2}\int t^{\frac{1}{2}}(1+\frac{2}{t^2})dt$
$I=\frac{1}{2}\int t^{\frac{1}{2}}dt+\int t^{-\frac{3}{2}}dt=\frac{1}{3}{t}^{\frac{3}{2}}-\frac{2}{\sqrt{t}}+c$

(C) $\int \frac{cos8x-cos7x}{1+2cos5x}dx$
$=\int \frac{-2sin\frac{5x}{2}sin\frac{x}{2}(3-2+2cos5x)dx}{1+2cos5x}$ $[\because sin3\left(\frac{5x}{2}\right)=3sin\frac{5x}{2}-4si{n}^3\frac{5x}{2}]$
$=\int (cos3x-cos2x)dx=\frac{sin3x}{3}-\frac{sin2x}{2}+c$

(D) $\int \frac{lnx}{x^3}dx=\left(lnx\right)(-\frac{1}{2x^2})+\int \frac{1}{x}.\frac{1}{2x^2}dx$
$=(-\frac{1}{2})\frac{lnx}{x^2}+(-\frac{1}{4})\frac{1}{x^2}+c$