New Notes Blog: `int dx/(x(sqrt(4x^2-1)))` Find the indefinite integral

Thursday, 20 August 2015

$\displaystyle\int\frac{{\left.{d}{x}}}{{{x}{\left(\sqrt{{{4}{{x}}^{{2}}-{1}}}\right)}}}$ Find the indefinite integral

We have to evaluate the integral : $\int \frac{dx}{x\sqrt{4x^2-1}}$

Let $\sqrt{4x^2-1}=u$

So, $\frac{1}{2\sqrt{4x^2-1}}.8x dx=du$

$\frac{4xdx}{\sqrt{4x^2-1}}=du$

$\frac{dx}{\sqrt{4x^2-1}}=\frac{du}{4x}$

Hence we have,

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

$=\int \frac{du}{u^2+1}$

$=tan^{-1}(u)+C$

$=tan^{-1}(\sqrt{4x^2-1})+C$ where C is a constant

We have to evaluate the integral : $\int \frac{dx}{x\sqrt{4x^2-1}}$

Let $\sqrt{4x^2-1}=u$

So, $\frac{1}{2\sqrt{4x^2-1}}.8x dx=du$

$\frac{4xdx}{\sqrt{4x^2-1}}=du$

$\frac{dx}{\sqrt{4x^2-1}}=\frac{du}{4x}$

Hence we have,

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

$=\int \frac{du}{u^2+1}$

$=tan^{-1}(u)+C$

$=tan^{-1}(\sqrt{4x^2-1})+C$ where C is a constant

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