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JEE Main 201912 AprilMorning Shift

Exam: JEE Main
Year: 2019
Date: 12 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2019 · 12 April · Morning Shift · Mathematics · Indefinite Integration

The integral ∫ 2 x ^ 3 - 1 x ^ 4 + x dx , is equal to

The integral

∫ \dfrac{2{x}^{3} - 1}{{x}^{4} + x}dx

, is equal to

Options

A $\dfrac{1}{2}\log_{e}\dfrac{{( {x}^{3} + 1 )}^{2}}{| {x}^{3} |} + C$
B $\log_{e}\dfrac{| {x}^{3} + 1 |}{{x}^{2}} + C$
C $\log_{e}\left|\dfrac{\left({x}^{3} + 1\right)}{x}\right| + C$
D $\dfrac{1}{2}\log_{e}\dfrac{| {x}^{3} + 1 |}{{x}^{2}} + C$

Correct answer

C. $\log_{e}\left|\dfrac{\left({x}^{3} + 1\right)}{x}\right| + C$

Step-by-step solution

$I = ∫ \dfrac{2{x}^{3} - 1}{{x}^{4} + x}dx$
$⇒ I = ∫ \dfrac{2x - \dfrac{1}{{x}^{2}}}{{x}^{2} + \dfrac{1}{x}}dx$
Put, ${x}^{2} + \dfrac{1}{x} = t$ $⇒ \left(2x - \dfrac{1}{{x}^{2}}\right)dx = dt$
$∴ I = ∫ \dfrac{dt}{t} = lo{g}_{e}\left(t\right) + C$
$= lo{g}_{e}\left({x}^{2} + \dfrac{1}{x}\right) + C$
$= lo{g}_{e}\left|\dfrac{\left({x}^{3} + 1\right)}{x}\right| + C$

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