Question

integration of 1/sqrt(-2x^2+3x+1)​

Answer 1

Step-by-step explanation:

Given te integrate

$\frac{1}{ \sqrt{ - 2 {x}^{2} + 3x + 1} }$

We have,

$\int \frac{dx}{ \sqrt{ - 2 {x}^{2} + 3x + 1 } }$

$= \frac{1}{ \sqrt{2} } \int \frac{dx}{ \sqrt{ - ({x}^{2} - \frac{3}{2}x - \frac{1}{2} ) } }$

$= \frac{1}{ \sqrt{2} } \int \frac{dx}{ \sqrt{ - ( {x}^{2} - \frac{3}{2}x + { (\frac{3}{4} )}^{2} - { (\frac{3}{4}) }^{2} - \frac{1}{2} }) }$

$= \frac{1}{ \sqrt{2} } \int \frac{dx}{ \sqrt{ - ((x - \frac{3}{4} )^{2} - \frac{1}{16} )} }$

$= \frac{1}{ \sqrt{2} } \int \frac{dx}{ \sqrt{ { (\frac{1}{4} )}^{2} - (x - \frac{3}{4} )^{2} } }$

$= \frac{1}{ \sqrt{2} } . \sin ^{ - 1} ( \frac{x - \frac{3}{4} }{ \frac{1}{4} } ) + c$

$= \frac{1}{ \sqrt{2} } \sin^{ - 1} (4x - 3) + c$