Cube root in definite integral I am having trouble solving this integral
$$\int_0^7\frac{{\rm d}x}{1+\sqrt[3]{x+1}}$$
I guess I have to take $x + 1$ as $t^3$, but it's can't really get the point.
 A: Let $x+1=(t-1)^{3}$. When $x=0$, $t=2$ and when $x=7$, $t=3$. We get
$$\int\frac{\text{d}x}{1+\sqrt[3]{x+1}}=\int\frac{3(t-1)^{2}\text{d}t}{t}=\int\left(\frac{3t^{2}}{t}-\frac{6t}{t}+\frac{3}{t}\right)\text{d}t=\frac{3t^{2}}{2}-6t+3\ln\vert t\vert+C$$
where $C\in\mathbb{R}$ is a constant. Which means:
\begin{align*}
\int_{0}^{7}\frac{\text{d}x}{1+\sqrt[3]{x+1}}&=\left[\frac{3t^{2}}{2}-6t+3\ln\vert t\vert\right]_{2}^{3}\\
&=\frac{27}{2}-18+3\ln(3)-6+12-3\ln(2)\\
&\approx 2.7164\end{align*}
which seems to be correct, according to WolframAlpha.
A: Hint. By the change of variable
$$
t=\sqrt[3]{x+1},\quad dx=3t^2dt,
$$ one gets
$$
\int_0^7\frac{dx}{1+\sqrt[3]{x+1}}=3\int_1^2\frac{t^2}{1+t}dt=3\int_1^2(t-1)\:dt+3\int_1^2\frac{1}{1+t}dt.
$$
A: The substitution proposed works well.  We have for $x+1=t^3$, $dx=3t^2\,dt$.  The integral of interest $I$ becomes
$$I=\int_{1}^{2}\frac{3t^2}{1+t}\,dt$$
Then, write the numerator as
$$3t^2=3(t+1)^2-6(t+1)+3$$
Can you finish?
A: Yes that is a good start.  Letting, as you say, $t^3= x+ 1$ then $\frac{1}{1+\sqrt[3]{x+ 1}}$ becomes $\frac{1}{1+ t}$.  To complete the problem, use the fact that $3t^2 dt= dx$.  Then the integral is $\int \frac{3t^2dt}{t+ 1}$.  Of course, you need to change the limits of integration from x to t: when x= 0, $t^3= 1$ so t= 1.  When x= 7, $t^3= 8$ so t= 2:
$\int_0^8 \frac{3t^2dt}{t+ 1}$.  You can simplify that by dividing t+ 1 into $3t^2$.
