Question

Find the anti derivative of x²/(2+3x³)³​

Answer 1

check the attachment_______♡

Answer 2

ANSWER:

$- \left( \dfrac{1 }{ 18 {(2 +3 {x}^{3} ) }^{2} } \right) + C$

GIVEN:

$\dfrac{x^{2}}{ (2+3x^{3} )^{3}}$

TO FIND:

Anti derivative of :

$\dfrac{x^{2}}{ (2+3x^{3} )^{3}}$

EXPLANATION:

$\displaystyle \int \frac{x^{2}}{ (2+3x^{3} )^{3}}dx$

$Let \ \ 2 + 3x^3 = t$

Differentiate with respect to t.

$\dfrac{d}{dx}(constant) = 0$

$\dfrac{d}{dx}( {x}^{3}) = 3 {x}^{2}$

$0 + 9 {x}^{2} \dfrac{dx}{dt} = 1$

${x}^{2} dx = \dfrac{dt}{9}$

Substitute value of (2 + 3x³) and x²dx

$\displaystyle \int \frac{1}{ (t)^{3}} \frac{dt}{9}$

$\displaystyle \dfrac{1}{9} \int \frac{1}{ (t)^{3}} dt$

$\displaystyle \dfrac{1}{9} \int (t)^{ - 3} dt$

$\displaystyle \int x^n = \frac{x^{n+1} }{n + 1} + C$

$\dfrac{1}{9} \left( \dfrac{ {(t)}^{ - 3 + 1} }{ - 3 + 1} \right) + C$

$\dfrac{1}{9} \left( \dfrac{ {(t)}^{ - 2} }{ - 2} \right) + C$

$- \left( \dfrac{1 }{ 18 {t}^{2} } \right) + C$

Substitute t = 2 + 3x³

$- \left( \dfrac{1 }{ 18 {(2 +3 {x}^{3} ) }^{2} } \right) + C$

$\displaystyle\int \frac{x^{2}}{ (2+3x^{3} )^{3}}dx = -\left( \dfrac{1 }{ 18 {(2 +3 {x}^{3} ) }^{2} } \right) + C$