Simple Definite Integral #1

Let us find the indefinite integral of $I = \large{\int \dfrac{x^3}{(x+1)^3 + 6e^x + 3x+5}dx = \ ?}$

Put $x+1 = u; dx=du$ , so

$I = \int \dfrac{(u-1)^3}{u^3 + 6e^{u-1} + 3(u-1)+5}dx$

$\Longrightarrow I = \int \dfrac{u^3 - 3u^2 + 3u - 1}{u^3 + 6e^{u-1} + 3u + 2}dx$

$\Longrightarrow I = \int \left(\dfrac{u^3 + 6e^{u-1} + 3u + 2}{u^3 + 6e^{u-1} + 3u + 2} - \dfrac{3u^2 + 6e^{u-1} + 3}{u^3 + 6e^{u-1} + 3u + 2} \right) dx$

$\Longrightarrow I = \int \left(1- \dfrac{(u^3 + 6e^{u-1} + 3u + 2)'}{u^3 + 6e^{u-1} + 3u + 2}\right) dx$

$\Longrightarrow I= u - \ln(u^3 + 3u + 6e^{u-1} +2) + C$

$\Longrightarrow I= x - \ln(x^3 + 3x^2 + 6x + 6e^x + 6) + C$

And substituting the limits gives our answer to be $\boxed{\ln \left(\dfrac{6e}{8+3e}\right)}$ .