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Question Number 35412 by mondodotto@gmail.com last updated on 18/May/18

$$\mathbf{evaluate}\int \sqrt{({\mathbf{t}}^2+1+\frac{3}{4}\mathbf{t})}\mathbf{dt}$$

Commented by prof Abdo imad last updated on 19/May/18

$$letputI=\int \sqrt{t^2+\frac{3}{4}t+1}dt$$$$I=\frac{1}{2}\int \sqrt{4t^2+3t+4}dtbut$$$$4t^2+3t+4={\left(2t\right)}^2+2\left(2t\right).\frac{3}{4}+\frac{9}{16}-\frac{9}{16}$$$$={(2t+\frac{3}{4})}^2-\frac{9}{16}changement2t+\frac{3}{4}=\frac{3}{4}ch\left(x\right)$$$$giveI=\frac{1}{2}\int \sqrt{\frac{9}{16}({ch}^{2}x-1)}\frac{3}{8}sh\left(x\right)dt$$$$=\frac{3}{16}.\frac{3}{4}\int {sh}^{2}xdt$$$$=\frac{9}{64}\int \frac{ch\left(2x\right)-1}{2}dt$$$$=\frac{9}{128}x-\frac{9}{128}\frac{1}{2}sh\left(2x\right)+\lambda butwehave$$$$ch\left(x\right)=\frac{4}{3}(2t+\frac{3}{4})=\frac{8t}{3}+1\Rightarrow x=argch(\frac{8t}{3}+1)$$$$=ln(\frac{8t}{3}+1+\sqrt{{(\frac{8t}{3}+1)}^2-1})$$$$sh\left(x\right)=\sqrt{{ch}^{2}x-1}=\sqrt{{(\frac{8t}{3}-1)}^2-1}and$$$$sh\left(2x\right)=2sh\left(x\right)ch\left(x\right)\Rightarrow$$$$I=\frac{9}{128}ln(\frac{8t}{3}+1+\sqrt{{(\frac{8t}{3}+1)}^2-1})$$$$-\frac{9}{128}(\frac{8t}{3}+1)\sqrt{{(\frac{8t}{3}-1)}^2-1}+\lambda$$$$$$

Commented by prof Abdo imad last updated on 19/May/18

$$sh\left(x\right)=\sqrt{{(\frac{8t}{3}+1)}^2-1}and$$$$I=\frac{9}{128}ln(\frac{8t}{3}+1+\sqrt{{(\frac{8t}{3}+1)}^2-1})$$$$-\frac{9}{128}(\frac{8t}{3}+1)\sqrt{{(\frac{8t}{3}+1)}^2-1}+\lambda$$

Answered by ajfour last updated on 18/May/18

$$\int \sqrt{{(t+\frac{3}{8})}^2+{\left(\frac{\sqrt{55}}{8}\right)}^2}dt$$$$=\left(\frac{8t+3}{16}\right)\sqrt{t^2+\frac{3t}{4}+1}+\frac{55}{128}\ln \mid t+\frac{3}{8}+\sqrt{t^2+\frac{3t}{4}+1}\mid +c.$$

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