Int(x/sqr(x^2+4x+8))dx
Not quite...
You should simplify the integral down to $\displaystyle 2\int \left(\sec\theta\tan\theta-\sec \theta\right)\,d\theta$ after applying the substitution. Keep in mind that $\displaystyle \int \sec\theta\,d\theta=\ln\left|\sec\theta+\tan\theta \right|+C$
Can you continue on?
Sure
Our integral was $\displaystyle \int\frac{x\,dx}{\sqrt{(x+2)^2+2^2}}$
Applying the trig substitution $\displaystyle x+2=2\tan\theta\implies x=2\left(\tan\theta-1\right)$, we see that $\displaystyle \,dx=2\sec^2\theta\,d\theta$.
Substituting all this into the integral, we have $\displaystyle \int\frac{2\left(\tan\theta-1\right)2\sec^2\theta\,d\theta}{\sqrt{4\left(\tan^ 2\theta+1\right)}}=4\int\frac{\left(\tan\theta-1\right)\sec^2\theta\,d\theta}{2\sec\theta}$ $\displaystyle =2\int\left(\tan\theta-1\right)\sec\theta\,d\theta=2\int\left(\sec\theta\ tan\theta-\sec\theta\right)\,d\theta$
Does this make sense?
Here's an alternate way of solving this problem. It might be easier or harder for you, depending on what you're used to.
Setup the integral as $\displaystyle \int {\frac{{x dx}}{{\sqrt {{{(x + 2)}^2} + 4} }}}
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Let $\displaystyle x + 2 = 2\sinh \theta$, since $\displaystyle {\sinh ^2}\theta + 1 = {\cosh ^2}\theta$
so $\displaystyle dx = 2\cosh \theta d\theta$ and $\displaystyle x = 2\sinh \theta - 2
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$\displaystyle \int {\frac{{(2\sinh \theta - 2)(2\cosh \theta )d\theta }}{{\sqrt {4{{\sinh }^2}\theta + 4} }}} = \frac{4}{2}\int {\frac{{\sinh \theta \cosh \theta - {{\cosh }}\theta }}{{\sqrt {{{\sinh }^2}\theta + 1} }} d\theta }$
$\displaystyle = 2\int {\frac{{\sinh \theta \cosh \theta - \cosh \theta }}{{\sqrt {{{\cosh }^2}\theta } }}} d\theta = 2\int {\frac{{\sinh \theta \cosh \theta }}{{\cosh \theta }} d\theta - 2\int {\frac{{\cosh \theta }}{{\cosh \theta }} d\theta } }
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$\displaystyle = 2\cosh \theta - 2\theta + c$
remember that $\displaystyle x + 2 = 2\sinh \theta $, so $\displaystyle \theta = {\sinh ^{ - 1}}\left( {\frac{{x + 2}}{2}} \right)
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use the identity $\displaystyle {\sinh ^2}\theta + 1 = {\cosh ^2}\theta
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$\displaystyle \cosh \theta = \sqrt {1 + {{\sinh }^2}\theta } = \sqrt {1 + {{\left( {\frac{{x + 2}}{2}} \right)}^2}} = \sqrt {\frac{{{x^2}}}{4} + x + 2} = \frac{1}{2}\sqrt {{x^2} + 4x + 8}
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and we have our answer:
$\displaystyle 2\cosh \theta - 2\theta + c = \sqrt {{x^2} + 4x + 8} - 2{\sinh ^{ - 1}}\left( {\frac{{x + 2}}{2}} \right) + c
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