If m is any natural number, then the value of the integral ∫ (x^ 3m + x^ 2m + x^ m ) (2x^ 2m + 3x^ m + 6) ^ (1 / m) dx is (where, C & is an arbitrary constant)

Correct answer: A. (1)/(6(m + 1)) \ 2x^ 3m + 3x^ 2m + 6x^ m \ ^ (1 / m) + 1 + C Put,& I = ∫ (x^ 3m + x^ 2m + x^ m )( (2x^ 3m + 3x^ 2m + 6x^ m )1 / m x )dx = ∫ (x^ 3m - 1 + x^ 2m - 1 + x^ m - 1 )(2x^ 3m + 3x^ 2m + 6x^ m )1 / mdx = (1)/(6m) ∫ (2x^ 3m + 3x^ 2m + 6x^ m )1 / m(6 mx ^ 3m - 1 + 6 mx ^ 2m - 1 + 6 mx ^ m - 1 )dx Now, let (2x^ 3m + 3x^ 2m + 6x^ m ) = t ⇒ (6 mx ^ 3m - 1 + 6 mx ^ 2m - 1 + 6 mx ^ m - 1 )dx = dt ⇒ I = (1)/(6m) ∫ t^ 1 / m dt = (1)/(6m) ((t (1 / m) + 1)/( (1 / m) + 1)) =(1)/(6(m + 1))(2x^ 3m + 3x^ 2m + 6x^ m )(1 / m) + 1 + C

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If m is any natural number, then the value of the integral ∫ (x^ 3m...

\text{m}

is any natural number, then the value of the integral

∫

\left(x^{3m} + x^{2m} + x^{m}\right){

\left(2x^{2m} + 3x^{m} + 6\right)}^{\left(1 / m\right)}\text{dx}

is (where,

C

& is an arbitrary constant)

Options

A $\dfrac{1}{6\left(m + 1\right)}{\left\{2x^{3m} + 3x^{2m} + 6x^{m}\right\}}^{\left(1 / m\right) + 1} + C$
B $\dfrac{1}{6m}{\left\{2x^{3m} + 3x^{2m} + 6x^{m}\right\}}^{\left(1 / m\right) + 1} + C$
C $\dfrac{1}{6m}{\left\{2x^{3m} + 3x^{2m} + 6x^{m}\right\}}^{1 / m} + C$
DNone of the above

Correct answer

A. $\dfrac{1}{6\left(m + 1\right)}{\left\{2x^{3m} + 3x^{2m} + 6x^{m}\right\}}^{\left(1 / m\right) + 1} + C$

Step-by-step solution

Put,& $I = ∫ \left(x^{3m} + x^{2m} + x^{m}\right)\left(\dfrac{\left(2x^{3m} + 3x^{2m} + 6x^{m}\right)1 / m}{x}\right)dx$

$= ∫ \left(x^{3m - 1} + x^{2m - 1} + x^{m - 1}\right)\left(2x^{3m} + 3x^{2m} + 6x^{m}\right)1 / mdx$

$= \dfrac{1}{6m} ∫ \left(2x^{3m} + 3x^{2m} + 6x^{m}\right)1 / m\left(6{mx}^{3m - 1} + 6{mx}^{2m - 1} + 6{mx}^{m - 1}\right)dx$

Now, let \left(2x^{3m} + 3x^{2m} + 6x^{m}\right) = t ⇒ \left(6{mx}^{3m - 1} + 6{mx}^{2m - 1} + 6{mx}^{m - 1}\right)dx = dt $</p><p>$ ⇒ I = \dfrac{1}{6m} ∫ t^{1 / m}dt = \dfrac{1}{6m} \left(\dfrac{t \left(1 / m\right) + 1}{ \left(1 / m\right) + 1}\right)

=\dfrac{1}{6\left(m + 1\right)}\left(2x^{3m} + 3x^{2m} + 6x^{m}\right)(1 / m) + 1 + C

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