Question

Integrate the functions
$\frac{x^2}{{(2+3x^3)}^3}$

• A. $\frac{1}{18(2+3x^3{)}^2}+c$
• B. $\frac{1}{9(2+3x^3{)}^2}+c$
• C. $-\frac{1}{18(2+3x^3{)}^2}+c$
• D. $-\frac{1}{9(2+3x^3{)}^2}+c$

Answer 1

Answer: (C)

$-\frac{1}{18(2+3x^3{)}^2}+c$

Let $I=\int \frac{x^2}{(2+3x^3{)}^3}dx$

Take $2+3x^3=t$

On differentiating, we have

$9x^{2}dx=dt$ $\Rightarrow x^{2}dx=\frac{dt}{9}$

$\Rightarrow I=\int \frac{dt}{9.t^3}$

$=\frac{1}{9}\times \frac{t^{-3+1}}{-3+1}+c$

$=\frac{1}{9}\times \frac{t^{-2}}{-2}+c$

$=-\frac{1}{18t^2}+c$

$=-\frac{1}{18(2+3x^3{)}^2}+c$