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Found in: Page 340

### Essential Calculus: Early Transcendentals

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

# 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

See the step by step solution

## 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=$$\sqrt 2 {c_1} - \sqrt 2 \ln (\sqrt {2)}$$