Approximation for integral involving a square root of a polynomial I am trying to get an expression for this integral
$$ \int_0^1 \mathrm{d}x \dfrac{x^6\log(1-x)} {\sqrt{\left( x^3 - \lambda(x-1)\right)^3}} $$
with $ \lambda \ll1$. 
I know this integral has no solution in terms of elementary functions so I am trying to see the behaviour in $\lambda$ when $\lambda \rightarrow 0$. I can't make a Taylor expansion for the square root in the denominator, because when $x \approx 0$, $\lambda$ and $x$ become comparable, and the expansion doesn't converge.
What I have tried is to split the integral in two intervals $[0,\lambda]$ and $[\lambda,1]$ and then expand for $x/\lambda \ll 1$ for the first interval and for $\lambda/ x \ll 1$ for the second one. However it seems I don't obtain the good behaviour. In fact, I'm not sure of the validity of this.
Any hint about how to get the good behaviour in $\lambda$? 
Thanks in advance!
 A: I am not sure I understand. When $\lambda=0$ the numerator is $-x^7$ since $\log(1-x)\sim -x$ and the denominator only $x^{9/2}$ so the integral converges. Can't you just use Lebesgue dominated convergence theorem? 
EDIT
OK, I understand now. Call
$$
g(\lambda)=\int_{0}^{1}\frac{x^{6}\log(1-x)}{(x^{3}-\lambda(x-1))^{3/2}}dx
$$
and
$$
f(x,\lambda)=\frac{x^{6}\log(1-x)}{(x^{3}-\lambda(x-1))^{3/2}}.
$$
Then
$$
\frac{\partial f}{\partial\lambda}(x,\lambda)=\frac{3}{2}\frac{(x-1)x^{6}%
\log(1-x)}{\left(  x^{3}+\lambda(1-x)\right)  ^{\frac{5}{2}}}%
$$
and so since $0\leq x^{3}\leq x^{3}+\lambda(1-x)$,
\begin{align*}
\left\vert \frac{\partial f}{\partial\lambda}(x,\lambda)\right\vert  &
=\frac{3}{2}\frac{(1-x)x^{6}|\log(1-x)|}{\left(  x^{3}+\lambda(1-x)\right)
^{\frac{5}{2}}}\leq\frac{3}{2}\frac{(1-x)x^{6}|\log(1-x)|}{x^{\frac{15}{2}}%
}\\
& \sim\frac{3}{2}x^{7-^{\frac{15}{2}}}=\frac{3}{2}x^{-^{\frac{1}{2}}}%
\end{align*}
near $x=0$, which is integrable (near $x=1$ there is no problem since $t\log t\to 0$), where we used the fact that $\log(1-x)\sim-x$
near $x=0$. So you can differentiate under the integral sign to find that
$$
g^{\prime}(0)=\int_{0}^{1}\frac{3}{2}\frac{(x-1)x^{6}\log(1-x)}{\left(
x^{3}\right)  ^{\frac{5}{2}}}dx
$$
and so$$
g(\lambda)=\int_{0}^{1}\frac{x^{6}\log(1-x)}{(x^{3})^{3/2}}dx+\lambda\int%
_{0}^{1}\frac{3}{2}\frac{(x-1)x^{6}\log(1-x)}{\left(  x^{3}\right)  ^{\frac
{5}{2}}}dx+o(\lambda)
$$
