Question

1/√5ˣ²-4,Integrate the given function defined on proper domain w.r.t. x.

Answer 1

Dear Student,

Answer:$\int\frac{1}{\sqrt{5x^{2} -4 } } dx=\frac{1}{\sqrt{5} } log(x+\sqrt{x^{2}-(\frac{2}{\sqrt{5} }) ^{2}  } +C$

Solution: ∫1/√5x²-4 dx =

formula used

$\int\frac{1}{\sqrt{x^{2} -a^{2} }  } dx=log(x+\sqrt{x^{2}-a^{2}  })+C$

so we have to convert the given function in the above mentioned form,by taking √5 common from denominator

$\int{\frac{1}{ \sqrt{5}( \sqrt{ {x}^{2} - ( { \frac{2}{ \sqrt{5} } })^{2} } }} dx$

here a = 2/√5

Apply formula

$\int\frac{1}{\sqrt{5} } \frac{1}{\sqrt{x^{2} -(\frac{2}{\sqrt{5} }) ^{2} } }dx\\ =\frac{1}{\sqrt{5} } log(x+\sqrt{x^{2}-(\frac{2}{\sqrt{5} } )^{2}  })+C$

or $=\frac{1}{\sqrt{5} } log(x+\sqrt{x^{2}-\frac{4}{5}  } +C\\ \\= \frac{1}{\sqrt{5} } log(x+\sqrt{\frac{5x^{2}-4 }{5} }+C$

Hope it helps you.

Thanks for asking such a nice question.

Answer 2

we have to find out $\int{\frac{1}{\sqrt{5x^2-4}}}\,dx$

$\int{\frac{1}{\sqrt{5x^2-4}}}\,dx$

=$\int{\frac{1}{\sqrt{5}}\frac{1}{\sqrt{x^2-4/5}}}\,dx$

= $\frac{1}{\sqrt{5}}\int{\frac{1}{\sqrt{x^2-\{\sqrt{\frac{4}{5}}\}^2}}}\,dx$

now use formula,
$\int{\frac{1}{\sqrt{x^2-a^2}}}\,dx=log|x+\sqrt{x^2-a^2}|+C$

= $\frac{1}{\sqrt{5}}log|x+\sqrt{x^2-\frac{4}{5}}|+C$

hence, answer is /1√5 log|x + √(x² - 4/5)| +C