3x Integral

$\large = \int_0^1 2x^3 cos(x^2)dx + \int_4^5 \frac{y+5}{y^2+y-2}dy$

$\color{#3D99F6} x^2 = u$

$\color{#3D99F6} 2xdx = du$

$\large = \int_0^1 ucos(u)du + \int_4^5 ( \frac{2}{y-1} - \frac{1}{y+2} ) dy$

$\large = usin(u) \Big |_0^1 - \int_0^1 cos(u)du + 2ln(y-1) \Big |_4^5 - ln(y+2) \Big |_4^5$

$\large = sin(1) + cos(u) \Big |_0^1 + 2ln(\frac43) - ln(\frac76)$

$\large = sin(1) + cos(1) - 1 + ln(\frac43 ^2 \cdot \frac67)$

$\large = ln(\frac{32}{21}) + sin(1) + cos(1) - 1$

So $\color{#D61F06} a=32, b=21, c=d=1, f=-1$

Then:

$\large \int_{21}^{32} (6z^2 + 6z - 6)dz$

$\large = 2z^3 + 3z^2 - 6z \Big |_{21}^{32}$

$\large = \color{#3D99F6} \fbox{48697}$