Question

integral of x^2tan^-1 (x^3)dx/1+x^6​

Answer 1

Answer:

$\bold{ \large{ \pink{ \boxed{ \boxed{ \dfrac{1}{2} \: {( {tan}^{ - 1} {x}^{3} )}^{2} + c}}}}}$

Step-by-step explanation:

$\displaystyle \int \dfrac{ {x}^{2} {tan}^{ - 1} {x}^{3} }{1 + {x}^{6} } dx \\ let \: \\ \: {tan}^{ - 1} {x}^{3} = t \\ differentiating \: with \: respect \: to \: x\: \\ \dfrac{1 \: (3 {x}^{2}) }{1 + { ({x}^{3}) }^{2} } dx \: = \: dt \\ \dfrac{ {x}^{2} }{1 + {x}^{6} } dx = \dfrac{dt}{3} \\ \displaystyle \int t \: dt \\ = \dfrac{ {t}^{1 + 1} }{1 + 1} + c \\ = \dfrac{ {t}^{2} }{2} + c \\ = \dfrac{1}{2} \: {( {tan}^{ - 1} {x}^{3} ) }^{2} + c$