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I search for a good approach (or a general solution technique) to integrate

$$\int_0^1 \frac{x}{\sqrt{a+bx+cx^2}} dx$$

My knowledge on integration ends with rational functions.

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2  
Hint: complete the square. –  1015 Jan 29 '13 at 18:23
1  
Hint the second: You will need the derivative of arcsin. –  Harald Hanche-Olsen Jan 29 '13 at 18:29
    
Thanks, that got me on the right track. –  shuhalo Jan 29 '13 at 18:38

2 Answers 2

We give an outline, but a complete outline. If we want to avoid complex numbers (and to some degree even if we don't) there are a number of cases, and these cases tend to have subcases. Each case can be handled without much difficulty. We do not end up with a universal formula. However, for each case you could follow the procedure described below and produce a formula for that case in terms of $a$, $b$, and $c$.

$1.$ Suppose $c=0$. The subcase $b=0$ is trivial. If $b\ne 0$ we can let $u=a+bx$.

If $c=ne 0$, the manipulation suggested at the beginning of the answer by juantheron is useful. After that manipulation, the top is replaced by a constant. From now on assume that manipulation has been done.

$2.$ Suppose $c\lt 0$. Complete the square in the expression $a+bx+c$. We get something like $k-(px+q)^2$, where in any interesting case $k$ is positive. The substitution $px+q=\sqrt{k}u$ transforms our definite integral into something of shape $\int\frac{a}{\sqrt{1-u^2}}\,du$, and we are in $\arcsin$ land.

$3.$ Suppose $c\gt 0$. Complete the square. Then our quadratic takes on shape $(px+q)^2 +k$ where $k$ is positive, negative, or $0$, three subcases.

The subcase $k=0$ is easy, just take the square root (carefully), we get $1$ over a linear expression, and we are in $\log$ land.

If $k$ is positive, let $px+q=\sqrt{k}\,u$. After some minor algebra, we want to find $\int \frac{du}{\sqrt{u^2+1}}$. This can be done in various ways, such as $u=\tan t$, or $u=\sinh t$. There are several other methods.

If $k$ is negative, let $px+q=\sqrt{-k}\,u$. soon we end up needing $\int \frac{du}{\sqrt{u^2-1}}$. Now again there are various tricks, such as $u=\sec t$, or $u=\cosh t$, other tricks.

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$\displaystyle \int\frac{x}{\sqrt{a+bx+cx^2}}dx = \frac{1}{2c}\int\frac{\left(2cx+b\right)-b}{\sqrt{a+bx+cx^2}}dx$

$ \displaystyle = \frac{1}{2c}\int\frac{2cx+b}{\sqrt{a+bx+cx^2}}dx-\frac{b}{2c}\int\frac{1}{\sqrt{a+bx+cx^2}}dx$

for First one put $a+bx+cx^2 = t^2$ and $\left(2cx+b \right)dx = 2tdt$

$ \displaystyle = \frac{1}{c}\sqrt{a+bx+cx^2}-\frac{b}{2c}.\frac{1}{\sqrt{c}}\int\frac{1}{\sqrt{x^2+\frac{b}{c}.x+\frac{a}{c}}}dx$

$\displaystyle = \frac{1}{c}\sqrt{a+bx+cx^2}-\frac{b}{2c\sqrt{c}}\int\frac{1}{\sqrt{\left(x+\frac{b}{2c}\right)^2+\left(\frac{\sqrt{b^2-4ac}}{2c}\right)^2}}dx$

$\displaystyle = \frac{1}{c}\sqrt{a+bx+cx^2}-\frac{b}{2c\sqrt{c}}\ln \left|\left(x+\frac{b}{2c}\right)+\sqrt{x^2+\frac{b}{c}.x+\frac{a}{c}}\right|+\mathbb{C}$

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That's a huge leap from the 2nd to last line to the last line. Care to fill in some details? –  Ron Gordon Jan 29 '13 at 20:36

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