_ ^ (1 - x^2 ) x ;dx = - Vidyadip

MCQ

$\int_{}^{} {(1 - {x^2})\log x\;dx = } $

✓
$\left( {x - \frac{{{x^3}}}{3}} \right)\log x - \left( {x - \frac{{{x^3}}}{9}} \right) + c$
B
$\left( {x - \frac{{{x^3}}}{3}} \right)\log x + \left( {x - \frac{{{x^3}}}{9}} \right) + c$
C
$\left( {x + \frac{{{x^3}}}{3}} \right)\log x + \left( {x + \frac{{{x^3}}}{9}} \right) + c$
D
None of these

✓

Answer

Correct option: A.

$\left( {x - \frac{{{x^3}}}{3}} \right)\log x - \left( {x - \frac{{{x^3}}}{9}} \right) + c$

a
(a)$\int_{}^{} {(1 - {x^2})\log x\,dx} = \int_{}^{} {\log x\,dx} - \int_{}^{} {{x^2}\log x\,dx} $
$ = x(\log x - 1) - \frac{{{x^3}\log x}}{3} + \frac{{{x^3}}}{9} + c$
$ = \left( {x - \frac{{{x^3}}}{3}} \right)\log x - \left( {x - \frac{{{x^3}}}{9}} \right) + c.$

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