Evaluating $\int^1_0\frac{x^2}{4x+5}dx$ How can I integrate this function ?
$$\int^1_0\frac{x^2}{4x+5}dx$$
 A: Divide the $x^2$ by $4x+5$, using ordinary division of polynomials. We get
$$\frac{x^2}{4x+5}=\frac{1}{4}x-\frac{5}{16}+\frac{25}{16}\frac{1}{4x+5}.$$
Now the integration should be straightforward.
Alternately, as a second choice, let $u=4x+5$.  Then $du=4\,dx$, so $dx=\frac{1}{4}du$. Also, $x=\frac{1}{4}(u-5)$, so $x^2=\frac{1}{16}(u^2-10u+25)$.  We end up with
$$\int_{u=5}^9 \frac{1}{64}\frac{u^2-10u+25}{u}\,du.$$
 The integration is easy, because of the cancellations.
A: Or, alternatively (same idea), 
$$\frac{x^2}{4x+5} = \frac{x}{4} + \frac{25}{64x + 80} - \frac{5}{16}$$
which you can probably integrate pretty readily.
A: $$I := \int^1_0 \frac{x^2}{4x+5} \ dx$$
Since the numerator is of a higher degree than the denominator, we need to use long division. Upon long division, we can rewrite our integral as
$$I = \int^1_0 \frac{x}{4} + \frac{25}{16(4x+5)} -\frac{5}{16} \ dx \tag{1}$$
Now simply integrate each piece and let $u = 4x+5, du = 4 \  dx$. Don't forget that
$$\int \frac{du}{u} = \ln  |u| + C$$
