If \int f(x) \dx=g(x), then \int x^5 f (x^3) \dx is equal to:

Correct Answer is : 13x3g(x3)−∫x2g(x3)dx+c

\frac{1}{3} x^3 g (x^3) -\int x^3 g (x^3) \dx+c

\frac{1}{3} x^3 g (x^3) -\int x^2 g (x^3) \dx+c

\frac{1}{3} x^3 g (x^3) -3 \int x^4 g (x^3) \dx+c

NIMCET 2023 Question Paper with solutions

Question : 38 of 120

Marks: +1, -0

∫ f(x) \dx=g(x)

, then

∫ x^5 f (x^3) \dx

is equal to:

Solution:

Let

x^3=

⇒ \;\; 3 x^2 \dx\;=d t

{\I}\;=∫ x^5 f (x^3) \dx

\;=∫ x^2 x^3 f (x^3) \dx

=\;{1}/{3} ∫ t f(t) \dt

Using integration by parts

{\I}\;=\;{1}/{3} [t ∫ f(t) \dt-∫ (\;{d t}/{d t} ∫ f(t) \dt) \dt]

\;=\;{1}/{3} [{\tg}(t)-∫ g(t) \dt]

\;=\;{1}/{3} {\tg}(t)-\;{1}/{3} ∫ g(t) \dt

x^3=t

and

\dt\;=3 x^2 \dx

\;=\;{1}/{3} x^3 g (x^3) -\;{3}/{3} ∫ x^2 g (x^3) \dx

\;=\;{1}/{3} x^3 g (x^3) -∫ x^2 g (x^3) \dx+c .

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