If I = ∫ (ln x)^ 5 x^ 2 + x^ 2 (ln x)^ 3 dx =k (ln x)^ 3 + 1 ((ln x)^ 3 - 2) + c (where c is the constant of integration), then 9k & is equal to

Correct answer: B. 2 Let 1 + (ln x) ^ 3 = t^ 2 ⇒ I = ∫ (t^ 2 - 1)2tdt 3t ⇒ I = (2)/(3)∫ (t^ 2 - 1)dt = 2t^ 3 9 - (2)/(3)t + c = (2)/(9)t(t^ 2 - 3) + c = (2)/(9) 1 + (ln x) ^ 3 ( (ln x) ^ 3 - 2) + c ∴ k = (2)/(9) ⇒ 9k = 2

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If I = ∫ (ln x)^ 5 x^ 2 + x^ 2 (ln x)^ 3 dx =k (ln x)^ 3 + 1 ((ln...

I = ∫ \dfrac{\left(ln x\right)^{5}}{x^{2} + x^{2}\left(ln x\right)^{3}}dx =k\sqrt{\left(ln x\right)^{3} + 1}\left(\left(ln x\right)^{3} - 2\right) + c

(where

c

is the constant of integration), then

9k

& is equal to

Options

A $4$
B $2$
C $6$
D $10$

Correct answer

B. $2$

Step-by-step solution

Let

1 + {\left(ln x\right)}^{3} = t^{2}

⇒ I = ∫ \dfrac{\left(t^{2} - 1\right)2tdt}{3t}

⇒ I = \dfrac{2}{3}∫ \left(t^{2} - 1\right)dt

= \dfrac{2t^{3}}{9} - \dfrac{2}{3}t + c

= \dfrac{2}{9}t\left(t^{2} - 3\right) + c

= \dfrac{2}{9}\sqrt{1 + {\left(ln x\right)}^{3}}\left({\left(ln x\right)}^{3} - 2\right) + c

∴ k = \dfrac{2}{9} ⇒ 9k = 2

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