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Integral Calculus (Definite Integral)

$\def\D{\displaystyle}$CIE 2020 winter paper 22 no 6

 Find the exact value of $\displaystyle\int_{2}^{4} \dfrac{(x+1)^{2}}{x^{2}} \mathrm{~d} x$.

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*********math solution*************

$\begin{array}{ll}\dfrac{(x+1)^{2}}{x^{2}}=\dfrac{x^{2}+2 x+1}{x^{2}}&=1+\dfrac{2}{x}+\dfrac{1}{x^{2}} \\\D\int_2^4\left( 1+\dfrac{2}{x}+\dfrac{1}{x^{2}}\right) \mathrm{~d} x&= \left[x+2 \ln x-\dfrac{1}{x}+(c)\right]_2^4 \\&=\left[4-2 \ln 4-\dfrac{1}{4}\right]-\left[2+2 \ln 2-\dfrac{1}{2}\right]\\&=\dfrac{9}{4}+2 \ln 2 \\\end{array}$


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