3
$\begingroup$

$\int \sqrt{x^2 + 2x}dx$

I have no clue what to do on this problem. It is in the trig substitution chapter so I know I have to use that somehow. I know that I can not complete the square because both terms are positive and will not give me a difference of squares.

I know u subsitution will not work because I get leftover x terms.

I know that I basically have to manipulate this problem algebraically before I can work with it but I just do not know how to do that.

I tried to factor out an x or -x but neither makes progress.

  • 1
    So it seems that I was wrong about the limit on the number of questions per month, see here: [Have the limits on number of questions per month/day been increased (or cancelled)?](http://meta.math.stackexchange.com/questions/4551/).2012-07-02

2 Answers 2

4

Same thing as the last one. Use $ \int \sqrt{x^2 + 2x }\ \ dx = \int \sqrt{x^2 + 2x + 1 - 1}\ \ dx = \int \sqrt{(x + 1)^2 - 1}\ \ dx $ and use the hint $\sec \theta = x+1$.

  • 4
    $dx$'s tsk tsk.2012-06-09
4

Ok. Here is a general method of solving integrals of the form $P(x) = ax^{2}+bx+c$.

\begin{align*} ax^{2}+bx+c &= a \cdot(x^{2}+\frac{b}{a} x) + c \\\ &= a \cdot \biggl(x^{2}+ \frac{b}{a} \cdot x +\frac{b^2}{4a^2}\biggr) + c - \frac{b^2}{4a} \\\ &= a \cdot \biggl(x+\frac{b}{2a}\biggr)^{2} + c-\frac{b^2}{4a} \end{align*}

Now after doing this put $\displaystyle x+\frac{b}{2a} = \frac{1}{\sqrt{a}} \cdot \sqrt{\biggl(c-\frac{b^2}{4a}\biggr)} \cdot \tan\theta$

  • 0
    Nice answer chandu +1.2012-06-09