Question

Integrate the question below
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Answer 1

It looks like the integral is

`\displaystyle \int_0^1 (2x - 8x^2)/(1+4x) \, dx`

Polynomial division yields

`(2x-8x^2)/(1+4x) = -2x + 1 - \frac1{1+4x}`

Now split up the integral as

`\displaystyle \int_0^1 \left(-2x + 1 - \frac1{1+4x}\right) \, dx = \int_0^1 (1-2x) \, dx - \int_0^1 (dx)/(1+4x)`

In the second integral, substitute
`u=1+4x` and
`du=4\,dx`. Then
`x=0 \implies u=1` and
`x=1 \implies u=5`, so

`\displaystyle \int_0^1 (dx)/(1+4x) = \frac14 \int_1^5 \frac{du}u`

and by the fundamental theorem of calculus, the integral we want evaluates to

`\displaystyle \int_0^1 \left(-2x + 1 - \frac1{1+4x}\right) \, dx = \int_0^1 (1-2x) \, dx - \frac14 \int_1^5 \frac{du}u \\\\ = (x - x^2)\bigg|_(x=0)^(x=1) - \frac14 \ln|u| \bigg|_(u=1)^(u=5) \\\\ = \bigg((1 - 1^2) - (0 - 0^2)\bigg) - \frac14 (\ln(5) - \ln(1)) = \boxed{-\frac{\ln(5)}4}`