Single Variable Calculus, Chapter 7, 7.7, Section 7.7, Problem 64

Find the integral $\displaystyle \int^1_0 \frac{1}{\sqrt{16 t^2 + 1}} dt$

If we let $u = 4t$, then $du = 4dt$, so $\displaystyle dt = \frac{du}{4}$. We know that the anti-derivative of $\displaystyle \frac{1}{\sqrt{u^2 + 1}}$ is $\sin h^{-1} x$. When $x = 0, u = 0$ and when $x = 1, u = 4$. Therefore,


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\begin{equation}
\begin{aligned}

\int^1_0 \frac{1}{\sqrt{16 t^2 + 1}} dt =& \int^4_0 \frac{1}{\sqrt{u^2 + 1}} \cdot \frac{du}{4}
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\int^1_0 \frac{1}{\sqrt{16 t^2 + 1}} dt =& \frac{1}{4} \int^4_0 \frac{1}{\sqrt{u^2 + 1}} du
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\int^1_0 \frac{1}{\sqrt{16 t^2 + 1}} dt =& \frac{1}{4} [\sin h^{-1} u]^4_0
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\int^1_0 \frac{1}{\sqrt{16 t^2 + 1}} dt =& \frac{1}{4} [\ln (u + \sqrt{u^2 + 1})]^4_0
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\int^1_0 \frac{1}{\sqrt{16 t^2 + 1}} dt =& \frac{1}{4} \left[ \ln (4 + \sqrt{(4)^2 + 1}) - \ln (0 + \sqrt{(0)^2 + 1}) \right]
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\int^1_0 \frac{1}{\sqrt{16 t^2 + 1}} dt =& \frac{1}{4} [\ln (4 + \sqrt{17}) - \ln (\sqrt{1})]
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\int^1_0 \frac{1}{\sqrt{16 t^2 + 1}} dt =& \frac{1}{4} [\ln (4 + \sqrt{17}) - \ln (1)]
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\int^1_0 \frac{1}{\sqrt{16 t^2 + 1}} dt =& \frac{1}{4} [\ln (4 + \sqrt{17}) - 0]
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\int^1_0 \frac{1}{\sqrt{16 t^2 + 1}} dt =& \frac{1}{4} \ln ( 4 + \sqrt{17})
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& \text{or}
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\int^1_0 \frac{1}{\sqrt{16 t^2 + 1}} dt =& \frac{\ln (4 + \sqrt{17})}{4}

\end{aligned}
\end{equation}
$