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i can't understand how to solve this issue: using an appropriate substitution, evaluate this integral: $$ \int \frac{1+x²}{\sqrt{1+x}}\mathrm{d}x $$ can any one solve this so i can understand how to do this.

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When you are facing a radical which has a linear sum in it, it works well to use that as the basis of the substitution. Here, you would take $ \ u = x + 1 \ $ , which will give you $ \ du = dx \ $ . To deal with the numerator, you need to solve your substitution equation for $ \ x \ $ , giving $ \ x = u - 1 \ $ . The integral becomes

$$ \int \ \frac{1+ x^2}{\sqrt{1 + x }} \ dx \ \rightarrow \ \int \frac{1 + (u - 1)^2}{\sqrt{u}} \ du \ . $$

You would then multiply out the polynomial in the numerator.

The point in doing this is that you now have a polynomial divided simply by the square root of the variable, which will leave you with a set of terms in the integrand which are all just fractional powers of $ \ u \ $ , something which is much easier to integrate.

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  • $\begingroup$ good idea to solve this! $\endgroup$ – Nimi Estela May 31 '13 at 17:43
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Let $u^2=1+x$. Then $2u\,du =dx$. Using the fact that $x=u^2-1$, we find that $x^2+1=u^4-2u^2+2$.

Substituting, we find that our integral is $$\int \frac{(u^4-2u^2+2)(2u)}{u}\,du.$$ There is cancellation, and we have reached $$\int (2u^4-4u^2+4)\,du$$ which is $$\frac{2}{5}u^5 -\frac{4}{3}u^3+4u+C.$$ Now replace $u$ by $(x+1)^{1/2}$.

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  • $\begingroup$ rock man... thank you so much :) $\endgroup$ – Nimi Estela May 31 '13 at 17:35
  • $\begingroup$ You are welcome. Note that this is a close relative of the answer by RecklessReckoner. Many would prefer that one. I like to delay dealing with fractional exponents. But for example if we wanted $\int \sin(5x+2)\,dx$, the right substitution would be $u=5x+2$. $\endgroup$ – André Nicolas May 31 '13 at 18:00
  • $\begingroup$ When the radical is only a square-root, it's about equally helpful to either use a substitution on the linear argument of the radical or use a "rationalizing" substitution. For cube-roots and "higher", rationalizing substitutions are the way to go... $\endgroup$ – colormegone May 31 '13 at 18:54

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