Expanding the integrand

Question

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!

Answer 1

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.

Answer 2

Hint:

$$\int(t-4)(t-2)^{4/5}dt=\int(t-2)^{9/5}-2(t-2)^{4/5}d(t-2)$$

Answer 3

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.