∫ee21(x+12)2+(152)2dxUsing the standard integral of tan, we have215tan−1(2x+115)∣ee2215[tan−1(2e2+115)−tan−1(2e+115)]{\int_e}^{e^2}\frac{1}{\sqrt{({x+\frac{1}{2}})^2+(\frac{\sqrt{15}}{2})^2}}dx\\ \text{Using the standard integral of tan, we have}\\ \frac{2}{\sqrt{15}}tan^{-1}(\frac{2x+1}{\sqrt{15}})|_e^{e^2}\\ \frac{2}{\sqrt{15}}[tan^{-1}(\frac{2e^2+1}{\sqrt{15}})-tan^{-1}(\frac{2e+1}{\sqrt{15}})]\\∫ee2(x+21)2+(215)21dxUsing the standard integral of tan, we have152tan−1(152x+1)∣ee2152[tan−1(152e2+1)−tan−1(152e+1)]
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