# How do you integrate int troot3(t-4)dt?

Jan 29, 2017

$\frac{3}{7} {\left(t - 4\right)}^{\frac{7}{3}} + 3 {\left(t - 4\right)}^{\frac{4}{3}} + C$

#### Explanation:

$I = \int t \sqrt[3]{t - 4} \textcolor{w h i t e}{.} \mathrm{dt}$

Apply the substitution $u = t - 4$. This also implies that $t = u + 4$ and $\mathrm{du} = \mathrm{dt}$. Then:

$I = \int \left(u + 4\right) \sqrt[3]{u} \textcolor{w h i t e}{.} \mathrm{du}$

We can write $\sqrt[3]{u}$ with a fractional exponent and then distribute:

$I = \int \left(u + 4\right) {u}^{\frac{1}{3}} \textcolor{w h i t e}{.} \mathrm{du}$

$I = \int {u}^{\frac{4}{3}} \textcolor{w h i t e}{.} \mathrm{du} + 4 \int {u}^{\frac{1}{3}} \textcolor{w h i t e}{.} \mathrm{du}$

Integrate both using the rule $\int {u}^{n} \textcolor{w h i t e}{.} \mathrm{du} = {u}^{n + 1} / \left(n + 1\right) + C$:

$I = {u}^{\frac{7}{3}} / \left(\frac{7}{3}\right) + 4 \left({u}^{\frac{4}{3}} / \left(\frac{4}{3}\right)\right) + C$

$I = \frac{3}{7} {u}^{\frac{7}{3}} + 3 {u}^{\frac{4}{3}} + C$

Since $u = t - 4$:

$I = \frac{3}{7} {\left(t - 4\right)}^{\frac{7}{3}} + 3 {\left(t - 4\right)}^{\frac{4}{3}} + C$