Class 12th

$\begin{array}{l}Let={\displaystyle {\int }_1^2\left(\frac{1}{x}-\frac{1}{2x^2}\right)}{e}^{2x}dx\\ Let,2x=t\Rightarrow dx=\frac{dt}{2}\\ When,x=1,t=2*1=2\\ x=2,t=2*2=4\end{array}$

$\begin{array}{l}So,I={\displaystyle {\int }_2^4\left(\frac{1}{\left(t/2\right)}-\frac{1}{t\left(\frac{t}{2}\right)}\right)}{e}^t\frac{dt}{2}\\ ={\displaystyle {\int }_2^4\left(\frac{2}{t}-\frac{2}{t^2}\right)}{e}^t\frac{dt}{2}\end{array}$

$={\displaystyle {\int }_2^4\left(\frac{1}{t}+\frac{\left(-1\right)}{t^2}\right)}{e}^{t}dt$ is in the form

$\begin{array}{l}Let,={\displaystyle {\int }_{-1}^1\frac{dx}{x^2+2x+5.}}\\ ={\displaystyle {\int }_{-1}^1\frac{dx}{x^2+2x+1+4}}\\ ={\displaystyle {\int }_{-1}^1\frac{dx}{{\left(x+1\right)}^2+4}}={\displaystyle {\int }_{-1}^1\frac{dx}{{\left(x+1\right)}^2+2^2.}}\end{array}$

Let x + 1 = t $\Rightarrow$ dx = dt

When x = 1, t = 1 + 1 = 2

x = –1, t = –1 + 1 = 0

$\begin{array}{l}I\; ={\displaystyle {\int }_0^2\frac{dt}{t^2+2^2}}\\ =\frac{1}{2}{\left[ta{n}^{-1}\frac{t}{2}\right]}_0^2\\ =\frac{1}{2}\left[ta{n}^{-1}\frac{2}{2}-ta{n}^{-1}\frac{0}{2}\right]\\ =\frac{1}{2}\left(ta{n}^{-1}1-ta{n}^{-1}0\right)\end{array}$

$=\frac{1}{2}[\pi /4$
$\begin{array}{l}\frac{\mathrm{}}{}-0]\\ =\pi /8\end{array}$

Kindly go through the solution

Kindly go through the solution

Kindly go through the solution

$\begin{array}{l}Let={\displaystyle {\int }_0^{1}si{n}^{-1}\left(\frac{2x}{1+x^2}\right)}dx\\ Putting,x=ta\mathrm{n\theta }\Rightarrow \theta =ta{n}^{-1}x\&dx=se{c}^2\mathrm{\theta d\theta }\\ When,x=0,\theta =ta{n}^{-1}\left(0\right)=0\end{array}$

Kindly go through the solution

$={\displaystyle {\int }_0^1\frac{x}{x^2+1}}dx=\frac{1}{2}{\displaystyle {\int }_0^1\frac{2x}{x^2+1}}dx$

Putting x² + 1 = t $\Rightarrow$ 2xdx = dt. So, that

When x = 0, t = 0² + 1 = 1

x = 1, t = 1² + 1 = 2

$\begin{array}{l}\therefore {\displaystyle {\int }_1^2\frac{dt}{t}}=\frac{1}{2}{\left[log1+1\right]}_1^2\\ =\frac{1}{2}\left[log2-log1\right]\\ =\frac{1}{2}log2\end{array}$

${\displaystyle {\int }_0^{2/3}\frac{dx}{4+9x^2}}equals={\displaystyle {\int }_0^{2/3}\frac{dx}{2^2+{\left(3x\right)}^2}}$

 $=\frac{1}{2}*\frac{{\left[ta{n}^{-1}\frac{3x}{2}\right]}_0^{2/3}}{3}\\ =\frac{1}{6}\left[ta{n}^{-1}\frac{3}{2}*\frac{2}{3}-ta{n}^{-1}\frac{3}{2}*0\right]\\ =\frac{1}{6}\left[ta{n}^{-1}\left(1\right)-tan\left(0\right)\right]\\ =\frac{1}{6}[\pi /4-0]\\ =\pi /24$

? Option (c) is correct.