Problem 31 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{t^{3}-2}{t+1} d t$$

Step-by-Step Solution

Answer

Question: Evaluate the integral of the rational function \(\int \frac{t^3 - 2}{t + 1} dt\). Answer: \(\frac{1}{3}t^3 - \frac{1}{2}t^2 + t + C\)

Step 1: Perform long division on the given function

We want to find the polynomial and remainder after dividing the numerator, \(t^3 - 2\), by the denominator, \(t + 1\). Doing this, we have: $$ t^3 - 2 = (t+1)(t^2-t+1) \\ $$ Now with the result, we rewrite our integral as follows: $$ \int \frac{t^3 - 2}{t + 1} dt = \int \frac{(t+1)(t^2 - t + 1)}{t + 1} dt \\ $$

Step 2: Simplify

Since the term \((t+1)\) cancels out, the expression becomes: $$ \int (t^2 - t + 1) dt $$

Step 3: Integrate term by term

Now, we can easily integrate this polynomial term by term: $$ \int (t^2 - t + 1) dt = \int t^2 dt - \int t dt + \int 1 dt $$

Step 4: Evaluation

Proceed to integrate each term: $$ \int t^2 dt - \int t dt + \int 1 dt = \frac{1}{3}t^3 - \frac{1}{2}t^2 + t + C $$ Thus, the solution to our integral is: $$ \int \frac{t^3 - 2}{t + 1} dt = \frac{1}{3}t^3 - \frac{1}{2}t^2 + t + C $$