If f ( x ) ;dx = ( x ), then x^5 ;f ( x^3 ) ;dx = - Vidyadip

MCQ

If $\smallint f\left( x \right)\;dx = \varphi \left( x \right)$, then $\smallint {x^5}\;f\left( {{x^3}} \right)\;dx = $

A
$\frac{1}{3}\left[ {{x^3}\varphi \left( {{x^3}} \right) - \smallint {x^2}\varphi \left( {{x^3}} \right)dx} \right] + c$
B
$\frac{1}{3}{x^3}\varphi \left( {{x^3}} \right) - 3\smallint {x^3}\varphi \left( {{x^3}} \right)dx + c$
✓
$\;\frac{1}{3}{x^3}\varphi \left( {{x^3}} \right) - \smallint {x^2}\varphi \left( {{x^3}} \right)dx + c$
D
$\;\frac{1}{3}\left[ {{x^3}\varphi \left( {{x^3}} \right) - \smallint {x^3}\varphi \left( {{x^3}} \right)dx} \right] + c$

✓

Answer

Correct option: C.

$\;\frac{1}{3}{x^3}\varphi \left( {{x^3}} \right) - \smallint {x^2}\varphi \left( {{x^3}} \right)dx + c$

c
$\int f(x) d x=\psi(x)$

$I=\int x^{5} f\left(x^{3}\right) d x$

$\text { put } x^{3}=t \quad$

$ \Rightarrow \quad x^{2} d x=\frac{d t}{3}$

$=\frac{1}{3} \int \mathrm{tf}(\mathrm{t}) \mathrm{dt}$

$=\frac{1}{3}\left[\mathfrak{t} \psi(\mathfrak{t})-\int \psi(\mathfrak{t}) \mathrm{d} \mathfrak{t}\right]$

$=\frac{1}{3}\left[x^{3} \psi\left(x^{3}\right)-3 \int x^{2} \psi\left(x^{3}\right) d x\right]+c$

$=\frac{1}{3} x^{3} \psi\left(x^{3}\right)-\int x^{2} \psi\left(x^{3}\right) d x+c$

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