### Video Transcript

Determine the indefinite integral of π₯ squared minus 19π₯ plus 70 over π₯ minus 14 with respect to π₯.

In order to determine this integral, we will first simplify the integrand. Note that 70 is divisible by 14, as 70 divided by 14 equals five. Therefore, π₯ squared minus 19π₯ plus 70 factorizes to π₯ minus 14 multiplied by π₯ minus five, as negative 14 multiplied by negative five equals 70, and negative 14 add negative five equals negative 19. So, the numerator of the integrand can be rewritten as π₯ minus 14 multiplied by π₯ minus five. We can cancel out the factor of π₯ minus 14 in the numerator and denominator to simplify the integrand to π₯ minus five. We are now required to compute the indefinite integral of π₯ minus five with respect to π₯.

In order to do this, recall that the integral of the sum or difference of two functions is the sum or difference of the integrals of the two functions. Therefore, the integral we are asked to evaluate can be rewritten as the integral of π₯ with respect to π₯ minus the integral of five with respect to π₯. Now, recall that in order to find the indefinite integral of the term π multiplied by π₯ to the πth power with respect to π₯, where π and π are constants such that π is not equal to negative one, we keep the coefficient π as it is, increase the exponent π by one, and divide by the new exponent. We then add the constant of integration π.

Therefore, in order to find the indefinite integral of π₯ with respect to π₯, we keep the coefficient one as it is, increase the exponent, one, by one to obtain a new exponent of two, and divide by the new exponent. We then add the constant of integration which we have denoted by π one. Letting π equal zero in the formula described earlier, we obtain that for all constants π, the indefinite integral of π with respect to π₯ is ππ₯ plus π, where π is the constant of integration. Therefore, the indefinite integral of the constant five with respect to π₯ is five π₯ plus π two, where π two denotes the constant of integration.

So, the integral given to us in the question evaluates to π₯ squared over two plus π one minus five π₯ plus π two. Removing the parentheses, we obtain π₯ squared over two plus π one minus five π₯ minus π two. We can collect together the constants π one and negative π two to form the constant π. So, the integral in question evaluates to π₯ squared over two minus five π₯ plus π, where π is the constant of integration.