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Compute the integral $$\int_1^e \sqrt{\frac{16}{t^2} + 4t^2 + 16}\ dt$$

I don't really have any good ideas myself as to how to approach the integral. Seems like an ordinary $u-$sub, i.e. taking $u$ to be the expression under the square root, does not work. It is supposed to be solvable by hand without too much trouble, so I'm assuming there is some clever or practical solution which I have not been able to find. Any suggestions?

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    $\begingroup$ Looks like the expression under the square root is the square of a nice function. You can factor it and cancel the square with the square root. $\endgroup$
    – user1266745
    Commented Dec 22, 2023 at 13:33

2 Answers 2

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Hint: $$\frac{16}{t^2} + 4t^2 + 16 =4\left(\left(\frac{2}{t}\right)^2 + t^2 + 2\cdot \frac{2}{t} \cdot t \right) = 4\left(\frac{2}{t} + t \right)^2 $$

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Since you are integrating for $t$ between $1$ and $e$ (so for positive values of $t$), you have that: \begin{align*} \int_1^e\sqrt{\frac{16}{t^2}+4t^2+16}\:\:dt&=2\int_1^e\sqrt{t^2+\frac{4}{t^2}+4}\:\:dt\\&=2\int_1^e\sqrt{\frac{t^4+4t^2+4}{t^2}}\:\:dt\\&=2\int_1^e\frac{t^2+2}{t}\:\:dt\\&=2\int_1^e\left(t+\frac{2}{t}\right)dt\\&=2\left\{\frac{t^2}{2}\bigg\rvert_1^e+2\ln(t)\bigg\rvert_1^e\right\}\\&=2\left\{\frac{e^2}{2}-\frac{1}{2}+2\right\}\\&=e^2+3 \end{align*}

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