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I started simplifying $$\int{\dfrac{x}{\sqrt{x^2+1}}dx}$$ but I always get this: $$\int{x(x^2+1)^{-1/2}dx}.$$

But I don't know how to follow by that way.

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4  
No matter how frustrating a mathematical problem can be, shit is inappropriate language here. –  user2468 Apr 25 '12 at 20:35
    
@J.D. Sorry. I didn't know about it. –  Garmen1778 Apr 25 '12 at 20:36
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Have you tried substituting $u = x^2 + 1$? –  user15464 Apr 25 '12 at 20:36
    
@ArturoMagidin Actually its not $$\int{x(x^2+1)^{-1}}$$. Its $$\int{x(x^2+1)^{-1/2}}$$. But thanks anyways. –  Garmen1778 Apr 25 '12 at 20:40

5 Answers 5

up vote 7 down vote accepted

Hint. Use the substitution $u=x^2+1$. Then $du = 2x\,dx$, so $x\,dx = \frac{1}{2}\,du$. Therefore, $$\frac{x}{\sqrt{x^2+1}}\,dx = \frac{\frac{1}{2}\,du}{\sqrt{u}} = \frac{1}{2}u^{-1/2}\,du.$$

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Here's a hint: $$ \int \frac{1}{\sqrt{1+x^2}}{\huge(}x\,dx{\huge)}. $$ If you don't know what that hints at, you should learn it. When you do understand what it hints at, you'll understand how to evaluate most integrals that are done by substitution.

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Wouldn't it be better to write it as $$\int {\frac{1}{{\sqrt {1 + {x^2}} }}d\left( {\frac{{{x^2}}}{2}} \right)} $$ to make it a hint?? –  Pedro Tamaroff Apr 25 '12 at 20:58
    
@PeterTamaroff : I don't think so. The point is to tell the student what to look for when the student sees the integral. –  Michael Hardy Apr 26 '12 at 12:51
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@PeterTamaroff ......and you're giving the free of charge something that they should be learning how to do by themselves. The point of the hint was to help them learn how to do that step. Your version does that step for them. –  Michael Hardy Apr 26 '12 at 15:01
    
I see. My idea was to point out to a substitution. I guess it'd have been more appropriate to put $$\int {\frac{1}{{\sqrt {1 + {x^2}} }}d\left( {\frac{{1 + {x^2}}}{2}} \right)} $$ –  Pedro Tamaroff Apr 26 '12 at 17:00

This looks like a problem for . . . u-substitution!

Let $u$ = $x^{2} + 1$. Thus, $du = 2x dx$.

So now, your integral is:

$$\frac{1}{2} \int \frac{1}{\sqrt{u}} du$$

$$\frac{1}{2} \cdot {2\sqrt{u}}$$

Substituting $u$ back in for terms of $x$ and multiplying the $\frac {1}{2}$ by $2$, you are left with

$$\sqrt{x^{2} + 1}$$

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Personally, I dislike the use of variable substitution, which is sort of mechanical, for problems that can be solved by applying concept. Not to mention that changing variables is always taken extremely lightly, as if we can just plug in any expression for $x$ and presto! new variable! For example $u = x^2+1$ is clearly not valid in all of $\mathbb R$ as it isn't injective, but it is in $\mathbb R^+$, which works out because that's all we need in this case, but no one seemed to care to use this justification! Also in for beginner calculus students, "$\mathrm dx$" means god knows what, all I need to know is that $\mathrm d(f(x))=f'(x)\mathrm dx,$ and hence we encourage liberal manipulation of meaningless symbols.

For these kinds of integrals just use the chain rule: $$ (f(u(x))' = f'(u(x))u'(x)\Rightarrow f(u(x))=\int f'(u(x))u'(x) \mathrm dx$$ So here just identify $u(x)=x^2+1$ and $u'(x)=2x$, so all we need is a factor of $2$ inside the integral which we can obtain if we also divide by $2$, which we can then factor out. I think this is a very good method in general, that is, the method of looking for when a function and its own derivative are present in an integral.

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Substituting $x^2 + 1 = y$, hence $xdx = \frac{1}{2}dy$. Therefore, the integral becomes $\int \frac{1}{2\sqrt{y}}dy$ which should be simple to evaluate.

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