Indefinite IntegrationHard
Question
f′(ax + b) {f(ax + b)}n dx is equal toOptions
A.
{f(ax + b)}n+1 + C, ∀ n except n = − 1
{f(ax + b)}n+1 + C, ∀ n except n = − 1B.
{f(ax + b)}n+1 + C, ∀ n
{f(ax + b)}n+1 + C, ∀ n C.
{f(ax + b)}n+1 + C, ∀ n except = − 1
{f(ax + b)}n+1 + C, ∀ n except = − 1D.{f(ax + b)}n+1 + C, ∀ n
{f(ax + b)}n+1 + C, ∀ n