If $\int \frac{d x}{x \sqrt{5 x^2-3}}=K \tan ^{-1} f(x)+C$ then

$f(x)=\sqrt{\frac{5}{3} x^2-1}, K=\frac{1}{\sqrt{5}}$

$f(x)=\frac{1}{2} \sqrt{5 x^2-3}, K=\frac{1}{\sqrt{5}}$

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\int \frac{d x}{x \sqrt{5 x^2-3}}=K \tan ^{-1} f(x)+C

then

View answer and explanation

Correct option: B

f(x)=\sqrt{\frac{5}{3} x^2-1}, K=\frac{1}{\sqrt{3}}

Option B (

f(x)=\sqrt{{\frac{{5}}{{3}} x^2-1}}, K=\frac{{1}}{{\sqrt{{3}}}}

) is correct because the integral can be solved by substitution, leading to an arctangent integral.