Ncert Solutions Chemistry Class 12th

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{dx}{1+cosx}}\\ ={\displaystyle \int \frac{dx}{2co{s}^{2}x/2}\left[?1+cosx=2co{s}^{2}x/2\right]}\\ =\frac{1}{2}{\displaystyle \int se{c}^2\frac{x}{2}dx}=\frac{1}{2}.2tan\frac{x}{2}+C=tan\frac{x}{2}+C\\ Hence,therequiredsolutionistan\frac{x}{2}+C.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{1+cosx}{x+sinx}}dx\\ Putx+sinx=t\Rightarrow \left(1+cosx\right)dx=dt\\ \therefore I={\displaystyle \int \frac{dt}{t}}=log\left|t\right|=log\left|x+sinx\right|+C\\ Hence,therequiredsolutionislog\left|x+sinx\right|+C.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{e^{6logx}-e^{5logx}}{e^{4logx}-e^{3logx}}}dx\\ \therefore I={\displaystyle \int \frac{e^{log{x}^6}-e^{log{x}^5}}{e^{log{x}^4}-e^{log{x}^3}}}dx\\ ={\displaystyle \int \frac{x^6-x^5}{x^4-x^3}dx}={\displaystyle \int \frac{x^2\left(x^4-x^3\right)}{x^4-x^3}dx}={\displaystyle \int x^{2}dx}\\ =\frac{1}{3}{x}^3+C\\ Hence,therequiredsolutionis\frac{1}{3}{x}^3+C.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{\left(x^2+2\right)}{x+1}}dx\\ \therefore I={\displaystyle \int \left[\left(x-1\right)+\frac{3}{x+1}\right]}dx\\ ={\displaystyle \int \left(x-1\right)dx}+3{\displaystyle \int \frac{1}{x+1}dx}\\ =\frac{x^2}{2}-x+3log\left|x+1\right|+C\\ Hence,therequiredsolutionis\frac{x^2}{2}-x+3log\left|x+1\right|+C.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}L.H.S.={\displaystyle \int \frac{2x+3}{x^2+3x}}dx\\ Put{x}^2+3x=t\\ \therefore \left(2x+3\right)dx=dt\\ \Rightarrow {\displaystyle \int \frac{dt}{t}}=log\left|t\right|\Rightarrow log\left|x^2+3x\right|+C=R.H.S.\\ L.H.S.=R.H.S.\\ Hence,proved.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}L.H.S.={\displaystyle \int \frac{2x-1}{2x+3}}dx\\ \Rightarrow {\displaystyle \int \left(1-\frac{4}{2x+3}\right)}dx\left[Dividingthenumeratorbythe\text{\hspace{0.17em}denominator}\right]\\ \Rightarrow {\displaystyle \int 1.}dx-4{\displaystyle \int \frac{1}{2x+3}dx}\Rightarrow {\displaystyle \int 1.}dx-\frac{4}{2}{\displaystyle \int \frac{1}{x+\frac{3}{2}}dx}\\ \Rightarrow {\displaystyle \int 1.}dx-2{\displaystyle \int \frac{1}{x+\frac{3}{2}}dx}\Rightarrow x-2log\left|x+\frac{3}{2}\right|+C\\ \Rightarrow x-2log\left|\frac{2x+3}{2}\right|+C\Rightarrow x-log\left|{\left(\frac{2x+3}{2}\right)}^2\right|+C\\ \left[?nlogm=log{m}^n\right]\\ \Rightarrow x-log\left|{\left(2x+3\right)}^2\right|-log{2}^2+C\\ \Rightarrow x-log\left|{\left(2x+3\right)}^2\right|+C_1=R.H.S.\left[where{C}_1=C-log{2}^2\right]\\ L.H.S.=R.H.S.\\ Hence,proved.\end{array}$