_ ^ ( x + 1 x ) ^3 dx = - Vidyadip

MCQ

$\int_{}^{} {{{\left( {x + \frac{1}{x}} \right)}^3}} dx = $

A
$\frac{1}{4}{\left( {x + \frac{1}{x}} \right)^4} + c$
✓
$\frac{{{x^4}}}{4} + \frac{{3{x^2}}}{2} + 3\log x - \frac{1}{{2{x^2}}} + c$
C
$\frac{{{x^4}}}{4} + \frac{{3{x^2}}}{2} + 3\log x + \frac{1}{{{x^2}}} + c$
D
None of these

✓

Answer

Correct option: B.

$\frac{{{x^4}}}{4} + \frac{{3{x^2}}}{2} + 3\log x - \frac{1}{{2{x^2}}} + c$

b
(b) $\int_{}^{} {{{\left( {x + \frac{1}{x}} \right)}^3}dx = \int_{}^{} {\left( {{x^3} + \frac{1}{{{x^3}}} + 3x + \frac{3}{x}} \right)\,dx} } $
$ = \frac{{{x^4}}}{4} - \frac{1}{{2{x^2}}} + \frac{{3{x^2}}}{2} + 3\log x + c$
$ = \frac{{{x^4}}}{4} + \frac{{3{x^2}}}{2} + 3\log x - \frac{1}{{2{x^2}}} + c.$

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