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Recently, I was given a question sheet with a lot of integrals, and I could solve all of them except for a particular type of them: $$I_1=\int \frac{dx}{(x+1)^2\sqrt{x^2+2x+2}}=\int \frac{dx}{x^2\sqrt{x^2+1}}=\frac{-\sqrt{x^2+1}}{x}+C$$ $$I_2=\int \frac{dx}{(x-1)^2\sqrt{x^2-x+1}}$$ $$I_3=\int \frac{dx}{(2x+1)\sqrt{x^2-8}}$$ $$I_4=\int \frac{dx}{x\sqrt{x^2+x+1}}=\log x + \log\left(x+2+2\sqrt{x^2+x+1}\right)+C$$

And the like($I_2,I_3$ also are expressible in elementary functions, but they are not that cute). The method I see that wolfram is using(trial pro subscription) is to complete the square inside the roots to transform them to $\sqrt{ax^2+b}$ and then substitute $x=\sqrt{\frac{-b}{a}}\sec x$ if $b$ is negative, and $x=\sqrt{\frac{b}{a}}\tan x$ if positive. Finally, it uses the tangent half-angle subsitution. I really don't think this is the best way.

In particular, the trigonometric substitutions in $I_1,I_4$ seem to be unnecessary, since at the end we transform back the resulting composition of trigonometric functions into a function that doesn't need trigonometry at all (e.g. $-\csc(\tan^{-1}x)=-\frac{\sqrt{x^2+1}}{x}$)

So my question is: Is there some easy way to deal with the integrals: $$\int\frac{dx}{(ax+b)\sqrt{x^2+c}}, \int\frac{dx}{(ax+b)^2\sqrt{x^2+c}}$$ Without having to do the secant-Weiertrass substitution?

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up vote 2 down vote accepted

for the type of $\displaystyle \int \frac{1}{(ax+b)^2 \sqrt{cx^2 + dx + e}}dx$, $\displaystyle \frac{1}{ax+b} = t$ transforms into $\displaystyle \frac{at}{\sqrt{bt^2 + ct + d}}$ which can be reduced into standard integrals.

The same seems to work in linear case too. $$\frac{1}{(ax+b) \sqrt{cx^2+ dx + e}} = \frac{at}{t \sqrt{bt^2+ct+d}} = \frac{a}{\sqrt{bt^2+ct+d}}$$

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Yeap, that seems to be quite efective! And in the linear case, some ideas? – chubakueno May 28 '14 at 3:24
@chubakueno same seems to work in linear case too. – Santosh Linkha May 28 '14 at 3:57
Hiper-facepalms Yes it clearly does. Thank you very much! I will wait a little more for answers and then accept yours :) – chubakueno May 28 '14 at 4:04
@chubakueno sure lol I didn't notice either :D – Santosh Linkha May 28 '14 at 4:05

Usually more direct, with $\sqrt {u^2 + 1},$ to take $u = \sinh t.$ With $\sqrt {u^2 - 1},$ to take $u = \cosh t.$ Tends to save several steps.

Relevant: it will still be necessary, to integrate a rational function of $\sinh x$ and $\cosh x,$ to use a stereographic projection. I worked it out: $$ v = \tanh \frac{x}{2} = \frac{e^x - 1}{e^x + 1}; \; \; \; \mbox{so} \; |v| < 1. $$ $$ x = 2 \operatorname{argtanh} x = \log \left( \frac{1+v}{1-v}\right) $$ $$ \sinh x = \frac{2v}{1-v^2} $$ $$ \cosh x = \frac{1+v^2}{1-v^2} $$ $$ dx = \frac{2}{1-v^2} \, dv $$

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I have not studied hyperbolic functions yet, but I am open to ideas. Will this work with the first case? – chubakueno May 28 '14 at 3:37
After a little bit of google and wolframalpha I discovered the usefulness of those functions. Thank you very much! – chubakueno May 28 '14 at 4:40

The Euler substitution $t=\sqrt{ax^2+bx+c}-\sqrt{a}x$ for positive $a$ will reduce all these integrals to rational functions.

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That would only work for integrals of the form $$\int\frac{dx}{\sqrt{ax^2+bx+c}}$$, i think. – chubakueno May 28 '14 at 3:47

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