How can I integrate this function ?
$\int^1_0\frac{x^2}{4x+5}dx$
How can I integrate this function ?
$\int^1_0\frac{x^2}{4x+5}dx$
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.
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.
$I := \int^1_0 \frac{x^2}{4x+5} \ dx$
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
$\int \frac{du}{u} = \ln |u| + C$