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$$\int \frac{x+3}{\sqrt{-x^2-4x+5}} \, dx$$

first $-x^2-4x+5=-(x+2)^2+9$

$$\int \frac{x+3}{\sqrt{9-(x+2)^2}}\, dx =\frac{1}{3}\int \frac{x+3}{\sqrt{1-(\frac{x+2}{3})^2}} \, dx$$

How should I continue from here?

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    $\begingroup$ There's an obvious substitution to make. $\endgroup$ – user296602 Jan 4 '16 at 19:19
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    $\begingroup$ A better way to do this is with partial fractions. $\endgroup$ – Arcturus Jan 4 '16 at 19:21
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    $\begingroup$ Factorise the denominator as $(1-x)(5+x)$ and use partial fractions. $\endgroup$ – Mark Bennet Jan 4 '16 at 19:23
  • $\begingroup$ Sorry all I forgot the square root $\endgroup$ – newhere Jan 4 '16 at 19:48
  • $\begingroup$ @newhere, I have deleted my no-longer-applicable answer. $\endgroup$ – Jose Arnaldo Bebita-Dris Jan 4 '16 at 19:50
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Put $x+2=3t$ and then the integral becomes $$\int{\frac{3t+1}{\sqrt{1-{{t}^{2}}}}\,dt}=\int{\frac{3t\,dt}{\sqrt{1-{{t}^{2}}}}}+\int{\frac{dt}{\sqrt{1-{{t}^{2}}}}}.$$

I'm sure you can take it from there.

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$$\frac{1}{3}\int \frac{x+3}{\sqrt{1-(\frac{x+2}{3})^2}} \, dx$$ Let $ u = (x+2)/3 $ and $ du = (1/3) dx$ then we have $$\int \frac{3u + 1}{\sqrt{1-(u)^2}} \, du.$$

Now what happens if you substitute $u = \sin(v)$?

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  • $\begingroup$ why substitute u=sin(v)? $\endgroup$ – newhere Jan 4 '16 at 21:03
  • $\begingroup$ Hint: $\cos^2(u) + \sin^2(u) = 1$ so $\cos^2(u) = \ldots$ $\endgroup$ – Chantry Cargill Jan 4 '16 at 21:05
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Arrange it so you have the derivative of the square-rooted expression on the top line, leaving a numerical remainder as the numerator of a second integral.

Note that $$\int\frac{f'(x)}{\sqrt{f(x)}}dx=2\sqrt{f(x)}$$ is useful here.

Therefore $$I=-\frac 12\int\frac{-2x-4}{\sqrt{-x^2-4x+5}}dx+\int\frac{1}{\sqrt{9-(x+2)^2}}dx$$ $$=-\sqrt{-x^2-4x+5}+\arcsin\left(\frac{x+2}{3}\right)+c$$

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