Ncert Solutions Maths class 12th

$\begin{array}{l}{\displaystyle {\int }_2^3\frac{dx}{x^2-1}}={\left[\frac{1}{2\times 1}log\frac{x-1}{x+1}\right]}_2^3\\ =\frac{1}{2}log\left|\frac{3-1}{3+1}\right|-\frac{1}{2}log\left|\frac{2-1}{2+1}\right|\end{array}$

$\begin{array}{l}=\frac{1}{2}log\frac{2}{4}-\frac{1}{2}log\frac{1}{3}\\ =\frac{1}{2}\left[log\frac{1}{2}-log\frac{1}{3}\right]\\ =\frac{1}{2}log\frac{1/2}{1/3}\\ =\frac{1}{2}log\frac{3}{2.}\end{array}$

$\begin{array}{l}{\displaystyle 1_{}^{}\frac{dx}{1+x^2}}={\left[ta{n}^{-1}x\right]}_0^1\\ =ta{n}^{-1}\left(1\right)-ta{n}^{-1}\left(0\right)\\ =\; \pi /4-0\\ =\; \pi /4\end{array}$

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

${\displaystyle {\int }_4^5{e}^{x}dx}={\left[e^x\right]}_4^5=e^5-e^4=e^4\left(e-1\right)$

$\begin{array}{l}{{\displaystyle {\int }_0^{\pi /2}cos2xdx=\left[\frac{sin2x}{2}\right]}}_0^{\pi /2}=\frac{sin2\times \pi /2}{2}-\frac{sin2\times 0}{2}\\ =\frac{si\mathrm{n\pi }}{2}-0\\ =0\end{array}$

$\begin{array}{l}{{\displaystyle {\int }_0^{/4}sin2xdx=\left[\frac{-cos2x}{2}\right]}}_0^{\pi /4}\\ =\left[\frac{-cos2\times \pi /4}{2}+\frac{cos2\times 0}{2}\right]\\ =\frac{-cos\pi /2}{2}+\frac{cos0}{2}\\ =0+\frac{1}{2}=\frac{1}{2}\end{array}$

$\begin{array}{l}{\displaystyle {\int }_1^2\left(4x^3-5x^2+6x+9\right)}dx.\\ ={\left[4\times \frac{x^4}{4}-5\times \frac{x^3}{3}+6\frac{x^2}{2}+9x\right]}_1^2\\ ={\left[x^4-\frac{5}{3}{x}^3+3x^2+9x\right]}_1^2\\ =\left[2^4-\frac{5}{x}\times 2^3+3\times 2^2+9\times 2\right]-\left[1^4-\frac{5}{3}\times 1^3+3\times 1^2+9\times 1\right]\\ =\left[16-\frac{40}{3}+12+18\right]-\left[1-\frac{5}{3}+3+9\right]\\ =\left[\frac{48-40+36+54}{3}\right]-\left[\frac{3-5+9+27}{3}\right]\\ =\frac{98}{3}-\frac{34}{3}=\frac{64}{3}.\end{array}$

$\begin{array}{l}{\displaystyle {\int }_2^3\frac{1}{x}dx}={\displaystyle {\int }_{-2}^3\frac{1}{x}dx}={\left[logx\right]}_2^3\\ =log3–log2\\ =log\frac{3}{2}\end{array}$