Maths Integrals

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

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

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

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

? Option (c) is correct.

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

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