Two Antiderivatives (VCE SSCE Mathematical Methods): Flashcards

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Two antiderivatives
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Value of rr where power rule fails

r=1r = -1

Why power rule fails when r=1r = -1

Division by zero

(ax+b)rdx\int (ax+b)^r \, dx when r1r \neq -1

1a(r+1)(ax+b)r+1+c\frac{1}{a(r+1)}(ax+b)^{r+1} + c

1ax+bdx\int \frac{1}{ax+b} \, dx

1alogeax+b+c\frac{1}{a}\log_e|ax+b| + c

First step in power rule for (ax+b)r(ax+b)^r

Increase power by 1

What to divide by in power rule

a(r+1)a(r+1)

What to divide by in logarithmic case

aa (coefficient of xx)

Why use absolute value in logeax+b\log_e|ax+b|

Handles both positive and negative cases

When integrating (ax+b)r(ax+b)^r, case requiring logarithm

r=1r = -1

Must include in general antiderivatives

Constant of integration cc

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