Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Evaluate the integral \(\int {\frac{{\sqrt {2{y^2} - 3} }}{{{y^2}}}} dy\)

\(\int {\frac{{\sqrt {2{y^2} - 3} }}{{{y^2}}}} .dy = ( - )\frac{{\sqrt {2{y^2} - 3} }}{y} + \sqrt 2 \ln |\sqrt 2 y + \sqrt {2{y^2} - 3|} \)+c

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1- To Find the integral

\(\begin{aligned}{l}\int {\frac{{\sqrt {2{y^2} - 3} }}{{{y^2}}}dy = \sqrt 2 \int {\frac{{\sqrt {{y^2} - \frac{3}{2}} }}{{{y^2}}}} } .dy\\\end{aligned}\)

\(\sqrt 2 \int {\frac{{\sqrt {{y^2} - ({{\sqrt {\frac{3}{2})} }^{^2}}} }}{{{y^2}}}} .dy\)

Step 2 -Formula used

\(\int {\frac{{\sqrt {{u^2} - {a^2}} }}{{{u^2}}} = ( - )\frac{{\sqrt {{u^2} - {a^2}} }}{u} + \ln |u + \sqrt {{u^2} - {a^2}} |} + c\)

=\(\begin{aligned}{l}\sqrt 2 \left( {\frac{{\sqrt {{y^2} - (\sqrt {\frac{3}{2}{)^2}} } }}{y}} \right) + \ln |y + \sqrt {{y^2} - (\sqrt {\frac{3}{2}{)^2}} |} + c\\\end{aligned}\)

=\(\frac{{( - )\sqrt {2{y^2} - 3} }}{y} + \sqrt 2 \ln |y + \sqrt {{y^2} - \frac{3}{2}|} + \sqrt 2 c\)

= \(( - )\frac{{\sqrt {2{y^2} - 3} }}{y} + \sqrt 2 (\ln |\sqrt {2y} + \sqrt {2{y^2} - 3} | - \ln \sqrt 2 ) + \sqrt 2 c\)

= \(\begin{aligned}{l}( - )\frac{{\sqrt {2{y^2} - 3} }}{y} + \sqrt 2 \ln |\sqrt 2 y + \sqrt {2{y^2} - 3|} + c\\\end{aligned}\)

Where c=\(\)\(\sqrt 2 {c_1} - \sqrt 2 \ln (\sqrt {2)} \)

Hence,\(\)\(\int {\frac{{\sqrt {2{y^2} - 3} }}{y}} .dy = \) \(\begin{aligned}{l}( - )\frac{{\sqrt {2{y^2} - 3} }}{y} + \sqrt 2 \ln |\sqrt 2 y + \sqrt {2{y^2} - 3|} + c\\\end{aligned}\)s

Where c=\(\)\(\sqrt 2 {c_1} - \sqrt 2 \ln (\sqrt {2)} \)

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Evaluate the integral \(\int {\frac{{\sqrt {2{y^2} - 3} }}{{{y^2}}}} dy\)

Step by Step Solution

Step 1- To Find the integral

Step 2 -Formula used

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Evaluate the integral \(\int {\frac{{\sqrt {2{y^2} - 3} }}{{{y^2}}}} dy\)

Step by Step Solution

Step 1- To Find the integral

Step 2 -Formula used

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