Question

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

Short answer

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

Step-by-step solution

Step 1- To Find the integral

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

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

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