What to do next with this indefinite integral?

Question

I'd like to evaluate the following indefinite integral:

$$\displaystyle\int \frac{dx}{(3+x^2)(\sqrt{1+x})}$$

I started by letting $y^2=1+x$ and, after simplifying, got here:

$$ 2\int\frac{dy}{y^4-2y^2+4}$$

I'm not sure what to do next. Does anyone have any advice?

score 3 · Answer 1 · answered Jun 19 '15 at 18:32

3

Factor the quartic by noticing it is a quadratic in disguise. Then seperate using partial fractions.

answered Jun 19 '15 at 18:32

RowanS

1,086

sorry ! will u please explain it more , how to factor that – Prasenjeet Symon Jun 19 '15 at 18:59
Let y^2= t. Now you have a quadratic expression in terms of t. After you have factorised, re-substitute. – Cataline Jun 19 '15 at 19:19

score 2 · Accepted Answer · answered Jun 19 '15 at 19:28

The factorization you need is \begin{align} y^4-2y^2+4&=y^4+4y^2+4-6y^2\\[4px] &=(y^2+2)^2-(\sqrt{3}\,y)^2\\[4px] &=(y^2-\sqrt{3}\,y+2)(y^2+\sqrt{3}\,y+2) \end{align} Then, with partial fractions, $$ \frac{1}{(y^2-\sqrt{3}\,y+2)(y^2+\sqrt{3}\,y+2)}= \frac{Ay+B}{y^2-\sqrt{3}\,y+2}+ \frac{Cy+D}{y^2+\sqrt{3}\,y+2} $$ and the integral becomes (almost) elementary.

score 1 · Answer 3 · answered Jun 19 '15 at 19:04

1

HINT: for the denominator we get $$3\left(\left(\frac{y^2-1}{\sqrt{3}}\right)^2+1\right)$$

answered Jun 19 '15 at 19:04

Dr. Sonnhard Graubner

95,283

What to do next with this indefinite integral?

3 Answers3