91Ó°ÊÓ

Divide the numerator \((t^3 - 2)\) by the denominator \((t + 1)\): Step (i) Divide the highest degree term of the numerator (\(t^3\)) by the highest degree term of the denominator (\(t\)) to get the first term of the quotient: \(t^2\). Step (ii) Multiply the first term of the quotient (\(t^2\)) by the denominator and subtract the result from the numerator: \begin{equation*} (t^3 - 2) - (t^2)(t + 1) = t^3 - 2 - (t^3 + t^2) = -t^2 - 2 \end{equation*} The remainder is now \(-t^2 - 2\). Since the remainder has a degree lower than the denominator, the division is complete. The result of the polynomial division is $$ \frac{t^3 - 2}{t + 1} \\ = t^2 - \frac{t^2 + 2}{t + 1} $$

Now, break down the given integral using the simplified fraction obtained in step 1: $$ \int \frac{t^3 - 2}{t + 1} d t \\ = \int \left(t^2 - \frac{t^2 + 2}{t + 1}\right) d t \\ = \int t^2 d t - \int \frac{t^2 + 2}{t + 1} d t $$

Integrate each term in the previous step: 1. The first integral is a simple power of t: $$ \int t^2 dt = \frac{1}{3}t^3 + C_1. $$ 2. The second integral is a rational function of t: $$ \int \frac{t^2 + 2}{t + 1} d t = \int \frac{t^2 + 2 - (t + 1) + (t + 1)}{t + 1} d t \\ = \int \left(\frac{-t^2 - t + 1}{t + 1} + 1\right) d t \\ = - \int \frac{t^2 + t - 1}{t + 1} d t + \int dt $$ First, integrate the second term: $$ \int dt = t + C_2 $$ The first term of the integral can be solved using substitution: let \(u = t + 1\), so \(d u = d t\), and \(t = u - 1\). Hence we get: $$ -\int \frac{(u - 1)^2 + (u - 1) - 1}{u} d u $$ Expanding and simplifying: $$ -\int \frac{u^2 - 2u + 1 + u - 1 - 1}{u} d u = -\int \frac{u^2 - u}{u} d u \\ = -\int (u - 1) du $$ The result is: $$ -\frac{1}{2}(u - 1)^2 + C_3 = -\frac{1}{2}(t + 1 - 1)^2 + C_3 = -\frac{1}{2}t^2 + C_3 $$

Combine the results of step 3 to find the final answer: \begin{equation*} \int \frac{t^3 - 2}{t + 1} d t \\ = \frac{1}{3}t^3 - \frac{1}{2}t^2 + t + (C_1 + C_2 + C_3) \\ = \frac{1}{3}t^3 - \frac{1}{2}t^2 + t + C \end{equation*} where \(C = C_1 + C_2 + C_3\) is the constant of integration.