Question

Integration.

Answer 1

?

Answer 2

\[\int\limits_{}^{} \frac{ 3x + 1 }{ 1 + (3x + 1)^4 }\]

Answer 3

Substitute 3x + 1 = u .. ?

Answer 4

With a dx at the end.

Answer 5

Doing that substitution, I get to \[\frac{ 1 }{ 3 }\int\limits_{}^{} \frac{ u }{ 1 + u^4 } du\]

Answer 6

And now it may seem silly, but I'm not sure where to go.

Answer 7

\(\Large u = (3x+1)^2\)

Answer 8

du = ... ?

Answer 9

remember that you have formula for
1/(x^2+a^2)

to convert 1+ X^4 into 1+U^2 , plugging in U =X^2 would makes sense, right ?

Answer 10

Use this formula here :

\[\int \left(\cfrac{1}{a^2 + x^2} \right) = \cfrac{1}{a} \tan^{-1} \left(\cfrac{x}{a}\right) \]

Answer 11

I missed dx there. in the LHS

Answer 12

and C in RHS. (+ C)

Answer 13

So then, using that, I get to a final answer of \[\frac{ 1 }{ 6} \tan^{-1} ((3x + 1)^2)\]

Answer 14

+ C

Answer 15

\(\huge \color{red}{\checkmark \quad \ddot \smile}\)

Answer 16

Well done.

Answer 17

Excellent, thanks I got it. ^-^