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Make a substitution to express the integrand as a rational function and then evaluate the integral

$$\int {\frac{2}{x\sqrt{x+1}}}\, dx$$

What is the substitution that I have to make?

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    $\begingroup$ Which bit of the integrand makes it not a rational function? What substitution could you make to eliminate this bit? NB at this stage don't worry whether or not the integral will be easy after the substitution, just make a substitution to get rid of the. . . whatever. $\endgroup$
    – David
    Feb 25, 2014 at 23:08
  • $\begingroup$ I understand that we can x^2 and multiply inside the root (1/sqrt(x^3+x^2)) but I dont get what to do after that... $\endgroup$
    – user96246
    Feb 25, 2014 at 23:13
  • $\begingroup$ That's not an integral substitution, that's just algebra and in this case only makes things worse. $\endgroup$
    – David
    Feb 25, 2014 at 23:21
  • $\begingroup$ Because I can't comment yet, is there any reason why the substitution of u = x + 1 can't be made for x in the denominator, and then derive u to obtain du = dx at which point solve for x ( x = u-1) and plug back in? This would yield: (1/(u-1)(u^(1/2))) du $\endgroup$ Feb 25, 2014 at 23:33
  • $\begingroup$ That would give you $(u-1)\sqrt u$ in the denominator. Not a rational function. And it really hasn't helped any. By setting $u = \sqrt{x + 1}$, squaring both sides to get $u^2 = x+1$, differentiating both sides: $2u\,du = dx$, so you can replace $dx$ with $2u\,du$, you have all you need for substituting to create a rational function. At any rate, it seems you haven't found my approach particularly helpful. $\endgroup$
    – amWhy
    Feb 25, 2014 at 23:36

2 Answers 2

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Let $u = \sqrt{x+1}$. Then $u^2 = x+1 \iff x = u^2 - 1$, and $dx = 2u\,du$. That gives you the integral

$$\int \dfrac{2u\,du}{u(u^2 - 1)} = 2\int \dfrac{du}{u^2 - 1}$$

Now we have our rational function. To evaluate, use the trigonometric substitution $u = \sec\theta$, and use the identity $$\tan^2 \theta + 1 = \sec^2\theta \iff \sec^2\theta - 1 = \tan^2\theta.$$

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  • $\begingroup$ Thank you! I just needed to know that substitution! $\endgroup$
    – user96246
    Feb 25, 2014 at 23:22
  • $\begingroup$ You're welcome! Actually: two substitutions! ;-) $\endgroup$
    – amWhy
    Feb 25, 2014 at 23:23
  • $\begingroup$ This type of substitution is known (for just this reason) as a "rationalizing substitution", and its usefulness is not limited to square-roots... $\endgroup$ Feb 25, 2014 at 23:39
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    $\begingroup$ Note that you don't need the trig substitution to solve this either; partial fractions works just fine (and arguably better) here. Since $u^2-1=(u-1)(u+1)$, look at $\dfrac{a}{u-1}+\dfrac{b}{u+1}$; multiply the 'a' term by $\frac{u+1}{u+1}$, and the 'b' term by $\frac{u-1}{u-1}$, then combine the terms and see how you can set $a$ and $b$ to get $\frac2{u^2-1}$ as the result. Once you've figured out how to write $\frac2{u^2-1} = \frac{a}{u-1}+\frac{b}{u+1}$ this way, you can integrate both of those in the usual way. $\endgroup$ Mar 4, 2014 at 17:49
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Your aim should be to get rid of the square root.

Warning: Spoiler below. :)

Try $u=\sqrt{x+1}$ so that $u^2-1=x$ Then $du = \frac{1}{2\sqrt{x+1}}\,dx = \frac{1}{2u}\,dx$ thus $2u\,du=dx$.

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