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Q8E

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Found in: Page 363

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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# Evaluate the integral$$\int {\frac{{{\rm{dx}}}}{{{{\rm{x}}^{\rm{2}}}\sqrt {{\rm{1 + }}{{\rm{x}}^{\rm{2}}}} }}}$$

Substitute trigonometrically$$\frac{{{\rm{dx}}}}{{{{\rm{x}}^{\rm{2}}}\sqrt {{\rm{1 + }}{{\rm{x}}^{\rm{2}}}} }}{\rm{ = - }}\frac{{\sqrt {{\rm{1 + }}{{\rm{x}}^{\rm{2}}}} }}{{\rm{x}}}{\rm{ + C}}$$

See the step by step solution

## Step1: We start with a trigonometric substitution$${\rm{x = tanu}}$$.

\begin{aligned}{c}\frac{{{\rm{dx}}}}{{{{\rm{x}}^{\rm{2}}}\sqrt {{\rm{1 + }}{{\rm{x}}^{\rm{2}}}} }}{\rm{x = tanu}}\\{\rm{dx = se}}{{\rm{c}}^{\rm{2}}}\\{\rm{ = }}\frac{{{\rm{se}}{{\rm{c}}^{\rm{2}}}{\rm{udu}}}}{{{\rm{ta}}{{\rm{n}}^{\rm{2}}}{\rm{u}}\sqrt {{\rm{1 + ta}}{{\rm{n}}^{\rm{2}}}{\rm{u}}} }}\end{aligned}

We currently use the term identity.

$${\rm{se}}{{\rm{c}}^{\rm{2}}}{\rm{u = 1 + ta}}{{\rm{n}}^{\rm{2}}}{\rm{u}}$$

\begin{aligned}{c}\frac{{{\rm{se}}{{\rm{c}}^{\rm{2}}}{\rm{udu}}}}{{{\rm{ta}}{{\rm{n}}^{\rm{2}}}{\rm{u}}\sqrt {{\rm{1 + ta}}{{\rm{n}}^{\rm{2}}}{\rm{u}}} }}{\rm{ = }}\frac{{{\rm{se}}{{\rm{c}}^{\rm{2}}}{\rm{udu}}}}{{{\rm{ta}}{{\rm{n}}^{\rm{2}}}{\rm{usecu}}}}\\{\rm{ = }}\frac{{{\rm{secudu}}}}{{{\rm{ta}}{{\rm{n}}^{\rm{2}}}{\rm{u}}}}\\{\rm{ = }}\frac{{\frac{{\rm{1}}}{{{\rm{co}}{{\rm{s}}^{\rm{2}}}}}}}{{\frac{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{u}}}}{{{\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{u}}}}}}{\rm{du}}\\{\rm{ = }}\frac{{{\rm{cosu}}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{u}}}}{\rm{du}}\end{aligned}

## Step2: We utilize substitution to calculate the final integral$${\rm{t = sinu}}$$.

\begin{aligned}{c}\frac{{{\rm{cosu}}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{u}}}}{\rm{du = t}}\\{\rm{ = sinu}}\end{aligned}

$${\rm{dt = cosudu}}$$

\begin{aligned}{c}{\rm{ = }}\frac{{{\rm{dt}}}}{{{{\rm{t}}^{\rm{2}}}}}\\{\rm{ = - }}\frac{{\rm{1}}}{{\rm{t}}}{\rm{ + C}}\\{\rm{ = - }}\frac{{\rm{1}}}{{{\rm{sinu}}}}{\rm{ + C}}\end{aligned}

## Step3: We must express our first substitution in order to return it $${\rm{sinu}}$$and$${\rm{x}}$$.

\begin{aligned}{c}{\rm{sinu = }}\frac{{{\rm{tanu}}}}{{\sqrt {{\rm{1 + ta}}{{\rm{n}}^{\rm{2}}}{\rm{u}}} }}\\{\rm{sinu = }}\frac{{\rm{x}}}{{\sqrt {{\rm{1 + }}{{\rm{x}}^{\rm{2}}}} }}\\\frac{{{\rm{dx}}}}{{{{\rm{x}}^{\rm{2}}}\sqrt {{\rm{1 + }}{{\rm{x}}^{\rm{2}}}} }}{\rm{ = - }}\frac{{\rm{1}}}{{{\rm{sinu}}}}{\rm{ + C}}\\{\rm{ = - }}\frac{{\sqrt {{\rm{1 + }}{{\rm{x}}^{\rm{2}}}} }}{{\rm{x}}}{\rm{ + C}}{\rm{.}}\end{aligned}

Conclusion finding a final solution$$\frac{{{\rm{dx}}}}{{{{\rm{x}}^{\rm{2}}}\sqrt {{\rm{1 + }}{{\rm{x}}^{\rm{2}}}} }}{\rm{ = - }}\frac{{\sqrt {{\rm{1 + }}{{\rm{x}}^{\rm{2}}}} }}{{\rm{x}}}{\rm{ + C}}{\rm{.}}$$

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