Maths Integrals

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

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

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

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

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

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Kindly go through the solution

Kindly go through the solution

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

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