Question

Integrate the question below​

Answer 1

It looks like the integral is

$\displaystyle \int_0^1 \frac{2x - 8x^2}{1+4x} \, dx$

Polynomial division yields

$\dfrac{2x-8x^2}{1+4x} = -2x + 1 - \dfrac1{1+4x}$

Now split up the integral as

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

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

$\displaystyle \int_0^1 \frac{dx}{1+4x} = \frac14 \int_1^5 \frac{du}u$

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

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