# Integral with substitution

1. Mar 22, 2006

### whkoh

Qn.

By using a suitable substituition, find
$$\int{\frac{1}{x\sqrt{1+x^n}}dx}$$

I haven't encountered this specific type of question before, so I went to use the obvious substitution
$$u^2=1+x^n$$, getting:

$$2u=n x^{n-1} \frac{dx}{du}\Leftrightarrow \frac{dx}{du}=\frac{2u}{n} x^{1-n}$$

Hence
$$\int{\frac{1}{x\sqrt{1+x^n}}dx}$$
$$=\int{\frac{2u}{n x^{1-n+1} \sqrt{u^2}}du$$
$$=\int{\frac{2ux^n}{nu}}du$$
$$=\int{\frac{2x^n}{n}}du$$
$$=\frac{2}{n}\int{x^n}du$$
$$=\frac{2}{n}\int{u^2-1}du$$
$$=\frac{2}{n}\left(\frac{1}{3}u^3 - u\right)+c$$
$$=\frac{2}{n}\left[\frac{1}{3}\left(1+x^n\right)^{\frac{3}{2}}-\sqrt{1+x^n}\right] +c$$

Is it correct?

2. Mar 22, 2006

### Hammie

$$=\int{\frac{2u}{n x^{1-n+1} \sqrt{u^2}}du$$

This is the first error I found.

x^(1-n) would be correct if it were in the numerator.

3. Mar 22, 2006

### TD

I've always found it a bad idea to 'mix' your variables while integrating (the original with the substitution). I think it's better to find expressions for your x's in function of the new variable and then replace all at once.

Letting $u^2 = 1 + x^n$, we have that:

$$x^n = u^2 - 1 \Leftrightarrow x = \left( {u^2 - 1} \right)^{1/n} \Leftrightarrow dx = \frac{{2u}}{n}\left( {u^2 - 1} \right)^{\left( {1 - n} \right)/n} du$$

Then substituting

$$\int {\frac{1}{{x\sqrt {1 + x^n } }}dx} \to \int {\frac{{2u}}{n}\frac{{\left( {u^2 - 1} \right)^{\left( {1 - n} \right)/n} }}{{\left( {u^2 - 1} \right)^{1/n} u}}du} = \frac{2}{n}\int {\frac{1}{{u^2 - 1}}du}$$

As you can see, this simplifies nicely to a classical integral.

I'll leave integration and substituting back for you