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SCREW L+S · Jun 10, 2022

Think I've screwed this up, can someone check
For some reason integral calc wasn't working as well

Thanks in advance

Run hard@thehsc · Jun 10, 2022

I think your u sub is wrong. Make it u = tan inverse x^3. Makes life way easier....

Lith_30 · Jun 10, 2022

I would recommend using the substitution $u=\tan^{-1}(x^3)$

$I=\int{\frac{6x^2}{1+x^6}\times\tan^{-1}(x^3)dx$

$\\let\ u=\tan^{-1}(x^3)\\\\du=\frac{3x^2}{1+x^6}dx$

$\\I=2\int{u}du\\\\I=u^2+c\\\\I=(\tan^{-1}(x^3))^2+c$

I can kinda see where you got the answer $\tan^{-2}(x^3)$ , but you must know that $\tan^{-1}(x)\neq\frac{1}{\tan(x)}$ , also your working out is kinda whacky and don't forget the +c in the end for non definite integrals.

SCREW L+S · Jun 11, 2022

Lith_30 said:
I would recommend using the substitution $u=\tan^{-1}(x^3)$

$I=\int{\frac{6x^2}{1+x^6}\times\tan^{-1}(x^3)dx$

$\\let\ u=\tan^{-1}(x^3)\\\\du=\frac{3x^2}{1+x^6}dx$

$\\I=2\int{u}du\\\\I=u^2+c\\\\I=(\tan^{-1}(x^3))^2+c$

I can kinda see where you got the answer $\tan^{-2}(x^3)$ , but you must know that $\tan^{-1}(x)\neq\frac{1}{\tan(x)}$ , also your working out is kinda whacky and don't forget the +c in the end for non definite integrals.

thanks bro

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SCREW L+S

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Run hard@thehsc

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Lith_30

o_o

SCREW L+S

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