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# How do you integrate int 1/sqrt(e^(2x)+12e^x+35)dx?

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#### Explanation

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#### Explanation:

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Steve M Share
May 18, 2018

$\int \setminus \frac{1}{\sqrt{{e}^{2 x} + 12 {e}^{x} + 35}} \setminus \mathrm{dx} = - \frac{1}{\sqrt{35}} \setminus \ln | \sqrt{{\left(35 {e}^{- x} + 6\right)}^{2} - 1} + 35 {e}^{- x} + 6 | + C$

#### Explanation:

We seek:

$I = \int \setminus \frac{1}{\sqrt{{e}^{2 x} + 12 {e}^{x} + 35}} \setminus \mathrm{dx}$

$\setminus \setminus = \int \setminus \frac{1}{\sqrt{{e}^{2 x} \left(1 + 12 {e}^{- x} + 35 {e}^{- 2 x}\right)}} \setminus \mathrm{dx}$

$\setminus \setminus = \int \setminus \frac{1}{{e}^{x} \sqrt{\left(1 + 12 {e}^{- x} + 35 {e}^{- 2 x}\right)}} \setminus \mathrm{dx}$

$\setminus \setminus = \int \setminus \frac{{e}^{- x}}{\sqrt{1 + 12 {e}^{- x} + 35 {e}^{- 2 x}}} \setminus \mathrm{dx}$

We can perform a substitution, Let:

$u = 35 {e}^{- x} + 6 \implies \frac{\mathrm{du}}{\mathrm{dx}} = - 35 {e}^{- x}$, and, ${e}^{- x} = \frac{u - 6}{35}$

The we can write the integral as:

$I = \int \setminus \frac{- \frac{1}{35}}{\sqrt{1 + 12 \left(\frac{u - 6}{35}\right) + 35 {\left(\frac{u - 6}{35}\right)}^{2}}} \setminus \mathrm{du}$

$\setminus \setminus = - \frac{1}{35} \setminus \int \setminus \frac{1}{\sqrt{1 + \frac{12}{35} \left(u - 6\right) + \frac{1}{35} {\left(u - 6\right)}^{2}}} \setminus \mathrm{du}$

$\setminus \setminus = - \frac{1}{35} \setminus \int \setminus \frac{1}{\sqrt{\frac{1}{35} \left(35 + 12 \left(u - 6\right) + {\left(u - 6\right)}^{2}\right)}} \setminus \mathrm{du}$

$\setminus \setminus = - \frac{1}{35} \setminus \int \setminus \frac{1}{\frac{1}{\sqrt{35}} \setminus \sqrt{35 + 12 u - 72 + {u}^{2} - 12 u + 36}} \setminus \mathrm{du}$

$\setminus \setminus = - \frac{1}{\sqrt{35}} \setminus \int \setminus \frac{1}{\sqrt{{u}^{2} - 1}} \setminus \mathrm{du}$

This is a standard integral, so we can write:

$I = - \frac{1}{\sqrt{35}} \setminus \ln | \sqrt{{u}^{2} - 1} + u | + C$

And if we restore the substitution, we get:

$I = - \frac{1}{\sqrt{35}} \setminus \ln | \sqrt{{\left(35 {e}^{- x} + 6\right)}^{2} - 1} + 35 {e}^{- x} + 6 | + C$

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