Question

17. f (ax b) [f(ax + b)]"

Answer 1

The given function is,

f( x )= f ′ ( ax+b ) ( f( ax+b ) ) n = 1 a f ′ ( ax+b )a ( f( ax+b ) ) n

Integrating both sides we get,

∫ f( x )dx = 1 a ∫ f ′ ( ax+b )a ( f( ax+b ) ) n dx (1)

Put f( ax+b )=t.

Differentiate both sides with respect to x,

f ′ ( ax+b ) d dx ( ax+b )= dt dx a f ′ ( ax+b )dx=dt (2)

From equation (1) and (2),

∫ f(t)dt = 1 a ∫ t n dt = t n+1 a( n+1 ) +c

Substitute value of t.

∫ f( x )dx = [ f( ax+b ) ] n+1 a( n+1 ) +c

Thus, integration of f( x ) is [ f( ax+b ) ] n+1 a( n+1 ) +c.