How do you find the antiderivative of #int 1/root3(1-5t) dt#?

Question

Emma Aldridge · Accepted Answer

I got: #-3/10root3((1-5t)^2)+c#

Have a look:

Ezekiel Alexander · Answer

#int(1/root3(1-5t))dt=-3/10(1-5t)^(2/3)+C#

#int(1/root3(1-5t))dt#

rewrite as

#int(1-5t)^(-1/3)dt#

now the outside of the bracket #=" constant "xx " bracket differentiated"#

so we are able to do this by inspection

using the power rule for integration ( add one to the power), and the fact that integration is the reverse of differentiation, let us try

#d/(dx)(1-5t)^(2/3)#

by the chain rule we get

#=2/3xx(-5)(1-5t)^(-1/3)=-10/3(1-5t)^(-1/3)#

so by comparing the integral and our 'inspected solution' we conclude

#int(1/root3(1-5t))dt=-3/10(1-5t)^(2/3)+C#

Axel Alvarez · Answer

undefined To find the antiderivative of $\int \frac{1}{\sqrt{3(1-5t)}} \, dt$, make the substitution $u = 3(1-5t)$. Then, differentiate both sides with respect to $t$ to get $du/dt = -15$, or $du = -15 dt$. Solving for $dt$, we have $dt = \frac{du}{-15}$.

Substitute back into the integral:

$$
\int \frac{1}{\sqrt{u}} \cdot \frac{du}{-15} = -\frac{1}{15} \int u^{-1/2} \, du
$$

Now, integrate $u^{-1/2}$ with respect to $u$:

$$
-\frac{1}{15} \left[ \frac{u^{1/2}}{1/2} \right] = -\frac{1}{15} \cdot 2\sqrt{u} = -\frac{2}{15}\sqrt{u}
$$

Substitute back for $u$:

$$
-\frac{2}{15}\sqrt{3(1-5t)} + C
$$

So, the antiderivative is $-\frac{2}{15}\sqrt{3(1-5t)} + C$, where $C$ is the constant of integration.