How do you integrate int 1/sqrt(e^(2x)+12e^x+32)dx using trigonometric substitution?

Mar 26, 2018

$- \frac{\sqrt{2}}{8} \ln | \frac{\left(16 + 3 {e}^{x}\right) + \sqrt{256 + 96 {e}^{x} + 8 {e}^{2 x}}}{e} ^ x | + C , \mathmr{and} ,$

$\frac{\sqrt{2}}{8} \left[x - \ln | \left\{\left(3 {e}^{x} + 16\right) + 2 \sqrt{2} \sqrt{{e}^{2 x} + 12 {e}^{x} + 32}\right\} |\right] + + C$.

Explanation:

We first use the substitution

${e}^{x} = t , \text{ so that, } {e}^{x} \mathrm{dx} = \mathrm{dt} , \mathmr{and} , \mathrm{dx} = \frac{\mathrm{dt}}{e} ^ x = \frac{\mathrm{dt}}{t}$.

$\therefore I = \int \frac{1}{\sqrt{{e}^{2 x} + 12 {e}^{x} + 32}} \mathrm{dx} = \int \frac{1}{t \sqrt{{t}^{2} + 12 t + 32}} \mathrm{dt}$.

Next, we let, $t = \frac{1}{y} . \therefore \mathrm{dt} = - \frac{1}{y} ^ 2 \mathrm{dy}$.

Now, $t \sqrt{{t}^{2} + 12 t + 32} = \frac{1}{y} \cdot \sqrt{\frac{1}{y} ^ 2 + \frac{12}{y} + 32}$,

$= \frac{\sqrt{1 + 12 y + 32 {y}^{2}}}{y} ^ 2$.

$\therefore I = \int {y}^{2} / \sqrt{1 + 12 y + 32 {y}^{2}} \cdot - \frac{1}{y} ^ 2 \mathrm{dy}$,

$= - \int \frac{1}{\sqrt{1 + 12 y + 32 {y}^{2}}} \mathrm{dy}$,

$= - \int \frac{\sqrt{8}}{\sqrt{8 \left(1 + 12 y + 32 {y}^{2}\right)}} \mathrm{dy}$,

$= - 2 \sqrt{2} \int \frac{1}{\sqrt{256 {y}^{2} + 96 y + 8}} \mathrm{dy}$,

$= - 2 \sqrt{2} \int \frac{1}{\sqrt{{\left(16 y + 3\right)}^{2} - 1}} \mathrm{dy}$,

$= - 2 \sqrt{2} \cdot \frac{1}{16} \ln | \left\{\left(16 y + 3\right) + \sqrt{256 {y}^{2} + 96 y + 8}\right\} |$,

$= - \frac{\sqrt{2}}{8} \ln | \left\{\left(\frac{16}{t} + 3\right) + \sqrt{\frac{256}{t} ^ 2 + \frac{96}{t} + 8}\right\} | \ldots \left[\because , y = \frac{1}{t}\right]$,

$= - \frac{\sqrt{2}}{8} \ln | \frac{\left(16 + 3 t\right) + \sqrt{256 + 96 t + 8 {t}^{2}}}{t} |$.

$\Rightarrow I = - \frac{\sqrt{2}}{8} \ln | \frac{\left(16 + 3 {e}^{x}\right) + \sqrt{256 + 96 {e}^{x} + 8 {e}^{2 x}}}{e} ^ x | + C , \mathmr{and} ,$

$I = \frac{\sqrt{2}}{8} \left[x - \ln | \left\{\left(3 {e}^{x} + 16\right) + 2 \sqrt{2} \sqrt{{e}^{2 x} + 12 {e}^{x} + 32}\right\} |\right] + + C$.

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