# How do you find the indefinite integral of int (4/(3t^2)+7/(2t))dt?

Feb 25, 2017

$\int \left(\frac{4}{3 {t}^{2}} + \frac{7}{2 t}\right) \mathrm{dt} = - \frac{4}{3 t} + \frac{7}{2} \ln t + c$

#### Explanation:

As $\int {t}^{n} \mathrm{dt} = {t}^{n + 1} / \left(n + 1\right) + c$, but for $n = - 1$, $\int {t}^{- 1} \mathrm{dt} = \int \frac{\mathrm{dt}}{t} = \ln t$

$\int \left(\frac{4}{3 {t}^{2}} + \frac{7}{2 t}\right) \mathrm{dt}$

= $\int \left(\frac{4}{3} {t}^{- 2} + \frac{7}{2} {t}^{- 1}\right) \mathrm{dt}$

= $\frac{4}{3} {t}^{- 1} / \left(- 1\right) + \frac{7}{2} \ln t + c$

= $- \frac{4}{3 t} + \frac{7}{2} \ln t + c$