Expanding the integrand Can anyone help me find the solution to this integral:
$$\int\limits{(t-4)(t-2)^{4/5}}dt?$$
I think I need to expand the integrand but I do not know how. Thanks a lot!
 A: Hint:
$$\int(t-4)(t-2)^{4/5}dt=\int(t-2)^{9/5}-2(t-2)^{4/5}d(t-2)$$
A: If you're still interested, you could use the following to solve integrals having the general form:
$$\int x^m (a+bx^n)^{r/s} dx$$
where $m,n,r$ and $s$ are integers and $n$ is positive.
To get to this form from your integral, we need to set $u=t-4$ which yields:
$$\int u(2+u)^{4/5} du$$
where $m=1, a=2,b=1,n=1,r=4$ and $s=5$.

1st case: if $\frac{m+1}{n}$ is an integer, then set $a+bx^n=w^s$

In your case $\frac{m+1}{n}=\frac{1+1}{1}=2$. Therefore, we set $$2+u=w^5$$ $$du=5w^4dw$$
And  obtain: $$\int(w^5-2)w^4(5w^4)dw$$
Integrating yields:
$$5(\frac{w^{14}}{14} -2\frac{w^9}{9})$$
Substituting $w$ then $u$, and factorizing yields:
$$ \frac{5}{126} (t-2)^{9/5}(9t - 46)$$
If you encounter a case where $\frac{m+1}{n}$ is not an integer then:

2nd case: If $\frac{m+1}{n} + \frac{r}{s}$ is an integer or is null, then set $a+bx^n=w^s x^n$.

Solving from here should be straightforward.
A: You can expand and use a substitution. The integrand can be expanded to give
\begin{equation*}
(t-2)^{4/5}t-4(t-2)^{4/5}
\end{equation*}
and you can do an integration term-by-term. For the first integral, use the substitution $u=t-2$ and use the substitution $s=t-2$ for the second integral. This gives 
$$\int u^{4/5}(u+2)du-4\int s^{4/5}ds\\
=\frac{10u^{9/5}}{9}+\frac{5u^{14/5}}{14}+\frac{10u^{9/5}}{9}+C$$
for a constant $C$. Substitute back to get the result. 
