JEE Main 2021 (18 March Morning Shift)

Question

The integral $\int {{{(2x - 1)\cos \sqrt {{{(2x - 1)}^2} + 5} } \over {\sqrt {4{x^2} - 4x + 6} }}} dx$ is equal to (where c is a constant of integration)

Accepted Answer

${1 \over 2}\sin \sqrt {{{(2x - 1)}^2} + 5} + c$

Answer

${1 \over 2}\cos \sqrt {{{(2x + 1)}^2} + 5} + c$

Answer

${1 \over 2}\cos \sqrt {{{(2x - 1)}^2} + 5} + c$

Answer

${1 \over 2}\sin \sqrt {{{(2x + 1)}^2} + 5} + c$

Aspire's Library

🎯 Crack NIMCET with Aspire Study

📂 Aspire Study Library

Aspire Study App

Aspire's Library

🎯 Crack NIMCET with Aspire Study

📂 Aspire Study Library

📲 Share This Question

Aspire Study App

Info.