Question

Find the integral of: x / sqrt(4x^2 + 6x) dx with limits of range increasing from -1/2e to 1/2e

Answer 1

Set $I = \int \frac {x}{\sqrt{(4x^{2} +6x)}} dx$

Completing the square in the denominator, we have that

$I = \int \frac {2x}{\sqrt{(4x + 3)^{2} -9}} dx$

Applying the substitution $4x +3 = 3 cosh \theta$

We have that

$I = \int \frac{2}{4} (\frac{3cosh\theta - 3}{\sqrt{9cosh^{2}\theta - 9}})(\frac{3}{4}sinh\theta d\theta)$

By simplifying, we have

$I = \frac{3}{8} \int (cosh\theta -1) d\theta$

Integrating term by term

$I = \frac{3}{8}(sinh\theta - \theta) + C$

Reverse the substitution

$I = \frac{3}{8}[(\frac{1}{3})\sqrt{(4x+3)^2-9} - \cosh^{-1}(\frac{4x + 3}{3})] + C$

Applying the upper and lower limits to the value of the definite integral

$\int_ {\frac{-1}{2e}}^{\frac{1}{2e}} \frac {x}{\sqrt{(4x^{2} +6x)}} dx$ is approximately $0.3504 + 0.7257i$

The reason why we have a complex solution is because of the lower limit of the integration.