The Student Room Group
Reply 1
The integral of (ax+b)^(n+1) = 1/a(n+1) * (ax + b)^(n+1) + C

Are you sure it's in your text book?
Shouldn't it be n instead of (n+1) in the LHS??
For the first one use the substitution u = ax+b, do du = adx (ie dx = du/a ), turning

Int [ (ax+b)n dx ] = (1/a) Int [ un du ] = (1/a) un+1/(n+1) = (1/a) (ax+b)n+1/(n+1)

For the second, let u = f(x), so du = f'(x)dx

Int [ f(x)nf'(x) dx ] = Int [ undu ] = un+1/(n+1) = f(x)n+1/(n+1)

+ constants etc

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