Question

Find the most general antiderivative of the function.

52. \(g\left( t \right) = \frac{{\left( {1 + t} \right)}}{{\sqrt t }}\)

Short answer

Thus, the most general antiderivative of given function is \(G\left( t \right) = 2\sqrt t - \frac{2}{{\sqrt t }} + C\).

Step-by-step solution

Simplify the given function

The given function is \(g\left( t \right) = \frac{{\left( {1 + t} \right)}}{{\sqrt t }}\). Simplify the function as follows.

\(\begin{aligned}{c}g\left( t \right) &= \frac{{\left( {1 + t} \right)}}{{\sqrt t }}\\ &= \frac{1}{{\sqrt t }} + \frac{t}{{\sqrt t }}\\ &= \frac{1}{{\sqrt t }} + \sqrt t \end{aligned}\)

Obtain antiderivative given function

Let\(G\left( t \right)\)is the antiderivative of\(g\left( t \right)\).

\(\begin{aligned}{c}g\left( t \right) &= \frac{1}{{\sqrt t }} + \sqrt t \\\frac{d}{{dt}}\left( {G\left( t \right)} \right) &= \frac{d}{{dt}}\left( {2\sqrt t } \right) + \frac{d}{{dt}}\left( {\frac{{ - 2}}{{\sqrt t }}} \right)\\G\left( t \right) &= 2\sqrt t - \frac{2}{{\sqrt t }} + C\end{aligned}\)

Hence, the most general antiderivative of given function is \(G\left( t \right) = 2\sqrt t - \frac{2}{{\sqrt t }} + C\).