# Evaluating $\int^1_0\frac{x^2}{4x+5}dx$

How can I integrate this function ?

$$\int^1_0\frac{x^2}{4x+5}dx$$

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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.

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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.

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$$I := \int^1_0 \frac{x^2}{4x+5} \ dx$$

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

$$\int \frac{du}{u} = \ln |u| + C$$

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