Expanding an Integrand

Question

I am trying to solve this integral but it looks like I do not get the right result. Can you please help me?

$$\int{(t-4)(t+2)^{\frac 4 5}}dt$$

I set $u=t+2$, so I get $\int{(u-6)u^{\frac 4 5}} \Bbb d t$ and then the solution I get is $$\frac 5 {14} (t+2)^{\frac {14} 5}-\frac{10} 3 (t+2)^{\frac 9 5} .$$

Can you please tell me where I am wrong?

Seems OK, remember when doing the substitution, you also have tu substitute the correct differential. In this case $dt=du$ so there is no problem. — Joaquin Liniado, Jun 09 '15 at 20:10
You can check by differentiation. Looks good, apart from the missing ${}+{}C$. — André Nicolas, Jun 09 '15 at 20:10

score 3 · Accepted Answer · answered Jun 09 '15 at 20:10

3

It looks fine. Just some cosmetic things. You indeed get $$\begin{align} \int (u-6)u^{4/5}\; d\color{red}{u} &= \int u^{9/5} -6u^{4/5}\; du \\ &= \frac{5}{14}u^{14/5} - 6\cdot\frac{5}{9}u^{9/5} + \color{red}{C} \\ &= \frac{5}{14}(t+2)^{14/5} - 6\cdot\frac{5}{9}(t+2)^{9/5} + \color{red}{C} \end{align} $$

answered Jun 09 '15 at 20:10

Thomas

43,555

ok then it was just a question of simplifying my solution... I used an online calculator to check the result and it was different to what I got thats why. Thanks. – ocram Jun 09 '15 at 20:13

3SAT · Answer 2 · 2015-06-09T20:26:13.470

$$\begin{align} \int{(t-4)(t+2)^{\frac {4}{5}}}dt &=\int({t(t+2)^{\frac {4}{5}}-4(t+2)^{4/5}})dt\\ &=\int{t(t+2)^{\frac {4}{5}}}dt-4\int{(t+2)^{4/5}}dt\\ \end{align}$$ Subtitute $u=t+2$ and $du=dt$ $$\begin{align} &=\int{(u-2)u^{4/5}}du-4\int{(t+2)^{4/5}}dt\\ &=\int{(u^{9/5}-2u^{4/5}})du-4\int{(t+2)^{4/5}}dt\\ &=\int{u^{9/5}}du-2\int{u^{4/5}}du-4\int{(t+2)^{4/5}}dt\\ &=\frac{5}{42}(t+2)^{9/5}(3t-22)+c \end{align} $$

Expanding an Integrand

2 Answers2