Value of a definite integral with irrational terms in the denominator Can someone help me to find the value of the following integration?
$$ \int_0^3\frac{dx}{\sqrt{x+1} + \sqrt{5x+1}} $$
I tried to rationalize the given expression, but that was not of much help.
I also tried to apply the following formula, 
$$ \int_0^a{f(x)dx} = \int_0^a{f(a-x)dx} $$ and got the following form 
$$ \int_0^3\frac{dx}{\sqrt{4-x} + \sqrt{16-5x}} $$
How do I reach to the final value?
 A: Sketch of solution:


*

*Rationalize denominators.

*Split the integral into two.

*One integral is $\int \frac{\sqrt{x+1}}{-4x} \; dx.$

*Substitute $u=\sqrt{x+1}$ to get $\frac{-1}{2}\int \frac{u^2}{u^2-1} \; du$.

*Polynomial long division and then either arctanh or partial fractions.

*Do the other integral similarly.
A: hint
If we multiply by the conjugate, 
it becomes
$$\int_0^3\frac {-\sqrt {x+1}}{4x}dx+\int_0^3\frac {\sqrt {5x+1}}{4x}dx $$
put $t=\sqrt {x+1} $ for the first and
$u=\sqrt {5x+1}$ for the second.
the first integral becomes
$$\int_1^2\frac {-2t^2}{4 (t^2-1)}dt $$
You can finish.
A: Rationalizing is probably the way to go. After rationalizing you can rewrite the integrand as,
$$ L =- \int_0^3 \frac{\sqrt{x+1} - \sqrt{5x+1}}{4x}\ dx $$
which can further be expanded into,
$$ L = -\frac 1 4 \int_0^3 \left( \frac{\sqrt{x+1}}{x} - \frac{\sqrt{5x+1}}{x} \right) \ dx  \quad\quad \text{(1)}$$
which has an antiderivative at limits $0, 3$,
$$L = -\frac 1 4 \left( -2\sqrt{x+1} + 2\sqrt{5x+1} + 2\tanh^{-1}\left(\sqrt{x+1}\right)-2\tanh^{-1}\left(\sqrt{5x+1}\right) \right)\big|_0^3 \quad \text{(2)}$$
you'll have to take the limit at $x=0$, but after evaluation you'll get 
$$ L = 1 - \frac 1 2 \log\left( \frac 5 3 \right) $$
Edit: We make the step from $\text{(1)}$ to $\text{(2)}$ due to the relation,
$$ \int\frac {\sqrt {a x + b}} {x}\, dx = 
 2\sqrt {a x + b} - 2\sqrt {b} \tanh^{-1}\left(\frac {\sqrt {a x + b}} {\sqrt {b}}\right) + c $$
