Question

Integrate this two questions ​

Answer 1

Simplify the integrands by polynomial division.

`(t^2)/(1 - 3t) = -\frac19 \left(3t + 1 - \frac1{1 - 3t}\right)`

`\frac t{1 + 4t} = \frac14 \left(1 - \frac1{1 + 4t}\right)`

Now computing the integrals is trivial.

5.

`\displaystyle \int (t^2)/(1 - 3t) \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - (\ln|1-3t|)/(27) + C}`

where we use the power rule,

`\displaystyle \int x^n \, dx = (x^(n+1))/(n+1) + C ~~~~ (n\\eq-1)`

and a substitution to integrate the last term,

`\displaystyle \int (dt)/(1-3t) = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)`

8.

`\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed)/(16) + C`

using the same approach as above.