Ncert Solutions Maths class 12th

Kindly go through the solution

Kindly go through the solution

$\begin{array}{l}{\displaystyle {\int }_0^2\frac{6x+3}{x^2+4}}dx={\displaystyle {\int }_0^2\frac{6x}{x^2+4}}dx+{\displaystyle {\int }_0^2\frac{3}{x^2+4}}dx\\ =3{\displaystyle {\int }_0^2\frac{2x}{x^2+4}}dx+3{\displaystyle {\int }_0^2\frac{1}{x^2+2^2}}dx\\ =3{\left[log\left|x^2+4\right|\right]}_0^2+\frac{3}{2}{\left[ta{n}^{-1}\frac{x}{2}\right]}_0^2\\ =3\left[log\left|2^2+4\right|-log\left|0^2+4\right|\right]+\frac{3}{2}\left[ta{n}^{-1}\frac{2}{2}-ta{n}^{-1}\frac{0}{2}\right]\\ =3\left[log\frac{8}{4}\right]+\frac{3}{2}[\pi /4-0]\\ =3log2+\frac{\mathrm{3\pi }}{}\end{array}$

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

$\begin{array}{l}{\displaystyle {\int }_0^{1}x{e}^{x^2}dx=Isay},\\ Putting,x^2=t\Rightarrow 2xdx=dt\to xdx=\frac{dt}{2}\\ ={\displaystyle {\int }_0^1{e}^t\frac{dt}{2}}=\frac{1}{2}{\displaystyle {\int }_0^1{e}^{t}dt}\end{array}$

When,

x = 0, t = 0

x = 1, t = 1

$\begin{array}{l}=\frac{1}{2}{\left[e^t\right]}_0^1\\ =\frac{1}{2}\left[e^1-e^0\right]=\frac{\left(e-1\right)}{2}\end{array}$

Kindly go through the solution

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

$\begin{array}{l}=\frac{1}{2}log\left|3^2+1\right|-\frac{1}{2}log\left|2^2+1\right|\\ =\frac{1}{2}\left[log\left|9+1\right|-log\left|4+1\right|\right]\\ =\frac{1}{2}log\frac{10}{5}\\ =\frac{1}{2}log2.\end{array}$

Kindly go through the solution