How do you integrate int 1 / (xsqrt(4x^2 +9))dx using trigonometric substitution?

Mar 21, 2018

$\int \frac{\mathrm{dx}}{x \sqrt{4 {x}^{2} + 9}} = \frac{1}{6} \ln \left\mid 3 - \sqrt{4 {x}^{2} + 9} \right\mid - \frac{1}{6} \ln \left(3 + \sqrt{4 {x}^{2} + 9}\right) + C$

Explanation:

Substitute:

$u = \sqrt{4 {x}^{2} + 9}$

$\mathrm{du} = \frac{4 x \mathrm{dx}}{\sqrt{4 {x}^{2} + 9}}$

Then:

$\int \frac{\mathrm{dx}}{x \sqrt{4 {x}^{2} + 9}} = \frac{1}{4} \int \frac{1}{x} ^ 2 \frac{4 x \mathrm{dx}}{\sqrt{4 {x}^{2} + 9}}$

and as:

${x}^{2} = \frac{{u}^{2} - 9}{4}$

then:

$\int \frac{\mathrm{dx}}{x \sqrt{4 {x}^{2} + 9}} = \frac{1}{4} \int \frac{4}{{u}^{2} - 9} \mathrm{du} = \int \frac{\mathrm{du}}{\left(u - 3\right) \left(u + 3\right)}$

Using now partial fractions decomposition:

$\frac{1}{\left(u - 3\right) \left(u + 3\right)} = \frac{A}{u - 3} + \frac{B}{u + 3}$

$1 = A \left(u + 3\right) + B \left(u - 3\right)$

$1 = \left(A + B\right) u + 3 \left(A - B\right)$

$\left\{\begin{matrix}A + B = 0 \\ A - B = \frac{1}{3}\end{matrix}\right.$

$\left\{\begin{matrix}A = \frac{1}{6} \\ B = - \frac{1}{6}\end{matrix}\right.$

So:

$\int \frac{\mathrm{dx}}{x \sqrt{4 {x}^{2} + 9}} = \frac{1}{6} \int \frac{\mathrm{du}}{u - 3} - \frac{1}{6} \int \frac{\mathrm{du}}{u + 3}$

$\int \frac{\mathrm{dx}}{x \sqrt{4 {x}^{2} + 9}} = \frac{1}{6} \ln \left\mid u - 3 \right\mid - \frac{1}{6} \ln \left\mid u + 3 \right\mid + C$

and undoing the substitution:

$\int \frac{\mathrm{dx}}{x \sqrt{4 {x}^{2} + 9}} = \frac{1}{6} \ln \left\mid 3 - \sqrt{4 {x}^{2} + 9} \right\mid - \frac{1}{6} \ln \left(3 + \sqrt{4 {x}^{2} + 9}\right) + C$